{"id":159,"date":"2023-11-08T16:10:09","date_gmt":"2023-11-08T16:10:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-the-greatest-common-factor\/"},"modified":"2026-02-13T18:24:53","modified_gmt":"2026-02-13T18:24:53","slug":"3-5-review-factoring","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/3-5-review-factoring\/","title":{"raw":"3.5 Review: Factoring","rendered":"3.5 Review: Factoring"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Review Objectives<\/h3>\r\n<ul>\r\n \t<li>Factor out the greatest common factor of a polynomial<\/li>\r\n \t<li>Factor by grouping<\/li>\r\n \t<li>Factor trinomials (when leading coefficient is one and when leading coefficient is not one)<\/li>\r\n \t<li>Factor perfect square trinomials and difference of squares<\/li>\r\n<\/ul>\r\n<\/div>\r\nClick on the following links to move to different types of factoring on this page.\r\n<a class=\"anchorLink\" href=\"#gcf\">Factor out the GCF<\/a>\r\n<a class=\"anchorLink\" href=\"#Grouping\">Factoring 4-term Polynomials by Grouping<\/a>\r\n<a class=\"anchorLink\" href=\"#Trinomial1\">Factoring Trinomials with Leading Coefficient of 1<\/a>\r\n<a class=\"anchorLink\" href=\"#Trinomialnot1\">Factoring trinomials with Leading Coefficient not 1<\/a>\r\n<a class=\"anchorLink\" href=\"#special\">Factoring Special Cases<\/a>\r\n<h2>Factoring Basics<\/h2>\r\n<strong>Factors<\/strong> are the building blocks of multiplication. They are the integers that you can multiply together to produce another integer: [latex]2[\/latex] and\u00a0[latex]10[\/latex] are factors of\u00a0[latex]20[\/latex], as are\u00a0[latex]4, 5, 1, 20[\/latex]. To factor an integer is to rewrite it as a product. [latex]20=4\\cdot{5}[\/latex] or [latex]20=1\\cdot{20}[\/latex]. In algebra, we use the word factor as both a noun \u2013 something being multiplied \u2013 and as a verb \u2013 the action of rewriting a sum or difference as a product.\u00a0<strong>Factoring<\/strong> is very helpful in simplifying expressions and solving equations involving\u00a0polynomials.\r\n\r\nThe <strong>greatest common factor<\/strong> (GCF) of two integers is the largest common factor to each integer. For instance, [latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest integer that divides evenly into both [latex]16[\/latex] and [latex]20[\/latex]. The GCF of polynomials works the same way: [latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest polynomial that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].\r\n\r\n<strong>When factoring a polynomial expression, our first step should be to check for a GCF.<\/strong> Look for the GCF of the coefficients, and then look for the GCF of the variables.\r\n<div class=\"textbox\">\r\n<h3>Greatest Common Factor<\/h3>\r\nThe <strong>greatest common factor<\/strong> (GCF) of a group of given polynomials is\u00a0the largest integer and highest degree of each variable that will divide evenly into each term of the polynomial.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFind the greatest common factor of [latex]25b^{3}[\/latex] and [latex]10b^{2}[\/latex].\r\n\r\n[reveal-answer q=\"210634\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"210634\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,25b^{3}=5\\cdot5\\cdot{b}\\cdot{b}\\cdot{b}\\\\\\,\\,10b^{2}=5\\cdot2\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5b^{2}\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the example above, the monomials have the factors\u00a0[latex]5[\/latex], [latex]b[\/latex], and [latex]b[\/latex] in common, which means their greatest common factor is [latex]5\\cdot{b}\\cdot{b}[\/latex], or simply [latex]5b^{2}[\/latex].\r\n\r\nThe video that follows gives an example of finding the greatest common factor of two monomials with only one variable.\r\n\r\nhttps:\/\/youtu.be\/EhkVBXRBC2s\r\n\r\nSometimes you may encounter a polynomial with more than one variable, so it is important to check whether both variables are part of the GCF. In the next example, we find the GCF of two terms which both contain two variables.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFind the greatest common factor of [latex]81c^{3}d[\/latex] and [latex]45c^{2}d^{2}[\/latex].\r\n[reveal-answer q=\"930504\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"930504\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,81c^{3}d=3\\cdot3\\cdot3\\cdot3\\cdot{c}\\cdot{c}\\cdot{c}\\cdot{d}\\\\45c^{2}d^{2}=3\\cdot3\\cdot5\\cdot{c}\\cdot{c}\\cdot{d}\\cdot{d}\\\\\\,\\,\\,\\,\\text{GCF}=3\\cdot3\\cdot{c}\\cdot{c}\\cdot{d}\\\\\\,\\,\\,\\,\\text{GCF}=9c^{2}d\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe video that follows shows another example of finding the greatest common factor of two monomials with more than one variable.\r\n\r\nhttps:\/\/youtu.be\/GfJvoIO3gKQ\r\n\r\nYou might be picking up on a useful shortcut for the variables. Notice that (assuming the variable is in common) we always select the smaller power of the variable.\r\n<h2 id=\"gcf\">Factoring out the Greatest Common Factor (GCF)<\/h2>\r\nNow that you have practiced identifying the GCF of terms with one and two variables, we can apply this idea to factoring\u00a0the GCF out of a polynomial. Notice that\u00a0the instructions are now \"Factor\" instead of \"Find the greatest common factor.\"\r\n\r\nTo factor a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property to rewrite the polynomial in factored form.\r\n<div class=\"textbox shaded\">\r\n<h3>Distributive Property Forward and Backward<\/h3>\r\nForward: Product of a number and a sum: [latex]a\\left(b+c\\right)=a\\cdot{b}+a\\cdot{c}[\/latex]. You can say that \u201c[latex]a[\/latex] is being distributed over [latex]b+c[\/latex].\u201d\r\n\r\nBackward: Sum of the products: [latex]a\\cdot{b}+a\\cdot{c}=a\\left(b+c\\right)[\/latex]. Here you can say that \u201c[latex]a[\/latex] is being factored out.\u201d\r\n\r\nWe first learned that we could distribute a factor over a sum or difference, now we are learning that we can \"undo\" the distributive property with factoring.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]25b^{3}+10b^{2}[\/latex].\r\n\r\n[reveal-answer q=\"716902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"716902\"]\r\n\r\nFind the GCF. From a previous example, you found the GCF of [latex]25b^{3}[\/latex] and [latex]10b^{2}[\/latex] to be [latex]5b^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,25b^{3}=5\\cdot5\\cdot{b}\\cdot{b}\\cdot{b}\\\\\\,\\,10b^{2}=5\\cdot2\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5\\cdot{b}\\cdot{b}=5b^{2}\\end{array}[\/latex]<\/p>\r\nRewrite each term with the GCF as one factor.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3} = 5b^{2}\\cdot5b\\\\10b^{2}=5b^{2}\\cdot2\\end{array}[\/latex]<\/p>\r\nRewrite the polynomial using the factored terms in place of the original terms.\r\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b\\right)+5b^{2}\\left(2\\right)[\/latex]<\/p>\r\nFactor out the [latex]5b^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b+2\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe factored form of the polynomial [latex]25b^{3}+10b^{2}[\/latex] is [latex]5b^{2}\\left(5b+2\\right)[\/latex]. You can check this by doing the multiplication. [latex]5b^{2}\\left(5b+2\\right)=25b^{3}+10b^{2}[\/latex].\r\n\r\nNote that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.\r\n\r\nFor example:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3}+10b^{2}=5\\left(5b^{3}+2b^{2}\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }5\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5b^{2}\\left(5b+2\\right) \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }b^{2}\\end{array}[\/latex]<\/p>\r\nNotice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.\r\n\r\nIn the following video, we show two more examples of how to find and factor the GCF from binomials.\r\n\r\nhttps:\/\/youtu.be\/25_f_mVab_4\r\n<div class=\"textbox\">\r\n<h3>How To: Given a polynomial expression, factor out the greatest common factor<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Identify the GCF of the coefficients.<\/li>\r\n \t<li>Identify the GCF of the variables.<\/li>\r\n \t<li>Write together to find the GCF of the expression.<\/li>\r\n \t<li>Determine what the GCF needs to be multiplied by to obtain each term in the expression.<\/li>\r\n \t<li>Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.<\/li>\r\n<\/ol>\r\n<\/div>\r\nWe will show a few examples of finding the GCF of a polynomial with several terms and sometimes with two variables. No matter how many terms the polynomial has, you can use the same technique described above to factor out its GCF.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor out the GCF. [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex]\r\n[reveal-answer q=\"112050\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"112050\"]\r\n\r\nFirst, find the GCF of the expression.\r\n\r\nThe GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. The GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions in the form [latex]{x}^{n}[\/latex] will always be the variable with the exponent of lowest degree.) And the GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. Put these together to find the GCF of the polynomial, [latex]3xy[\/latex].\r\n\r\nNext, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that [latex]3xy\\left(2{x}^{2}{y}^{2}\\right)=6{x}^{3}{y}^{3},3xy\\left(15xy\\right)=45{x}^{2}{y}^{2}[\/latex], and [latex]3xy\\left(7\\right)=21xy[\/latex].\r\n\r\nFinally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by:\r\n\r\n[latex]3xy\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]\r\n\r\nAfter factoring, we can check our work by multiplying. Use the distributive property to check.\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will see two more\u00a0example of how to find and factor out the greatest common factor of a polynomial.\r\n\r\nhttps:\/\/youtu.be\/3f1RFTIw2Ng\r\n\r\nIn the next example, the leading coefficient is negative. In this case, it is convention to include the negative with the GCF. In addition, we will see that this can assist us in other ways as well.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor out the GCF. [latex]-24x^8+32x^5[\/latex]\r\n\r\n[reveal-answer q=\"707771\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"707771\"]\r\n\r\nThe GCF of [latex]24[\/latex] and [latex]32[\/latex] is 8, while the GCF of [latex]x^8[\/latex] and [latex]x^5[\/latex] is [latex]x^5[\/latex]. However,\u00a0because the leading coefficient is negative, we will include the negative in the overall GCF.\r\n<p style=\"text-align: center;\">[latex]GCF = -8x^5[\/latex]<\/p>\r\nNext, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that\r\n<p style=\"text-align: center;\">[latex]-24x^8=-8x^5(3x^3)[\/latex] and [latex]32x^5=-8x^5(-4)[\/latex]<\/p>\r\nThus, factoring out the GCF gives us,\r\n<p style=\"text-align: center;\">[latex]-8x^5(3{x}^{3}-4)[\/latex]<\/p>\r\nNote that the effect of factoring out a negative term is that all signs of the remaining polynomial will be the opposite of the original terms.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAnother interesting situation we may encounter is where the GCF could even include an entire expression containing multiple terms, as shown in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor out the GCF. [latex]3x^2(2x-5)+7(2x-5)[\/latex]\r\n\r\n[reveal-answer q=\"497376\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"497376\"]\r\n\r\nResisting the temptation to distribute, we can view this as a polynomial with two terms: [latex]3x^2(2x-5)[\/latex] and [latex]7(2x-5)[\/latex]. The GCF of [latex]3[\/latex] and [latex]7[\/latex] is simply [latex]1[\/latex] and the terms do not share any power of [latex]x[\/latex] in common. However, each term contains the expression [latex](2x-5)[\/latex], so this will be the GCF.\r\n<p style=\"text-align: center;\">[latex]GCF=(2x-5)[\/latex]<\/p>\r\nWe can factor this out using the same strategy we used on all previous problems, writing the expression as a product of the GCF with a sum comprised of what we would need to multiply by to obtain each original term. So, factoring out the GCF yields.\r\n<p style=\"text-align: center;\">[latex](2x-5)(3{x}^{2}+7)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2 id=\"Grouping\">Factor by Grouping<\/h2>\r\nFactoring by grouping is a technique that allows us to factor a polynomial whose terms don\u2019t all share a GCF. In the following example, we will introduce you to the technique. Remember, some of the main reasons we want to be able to factor well \u00a0is because it will help solve polynomial equations and graph quadratic functions more easily.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]a^2+3a+5a+15[\/latex]\r\n\r\n[reveal-answer q=\"835783\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"835783\"]\r\n\r\nNotice that there is no GCF other than 1 for all four terms. Resist the temptation to combine the like terms, as instead we are going to group the problem into two pairs of terms and then search for the GCF of each.\r\n<p style=\"text-align: center;\">[latex](a^2+3a)+(5a+15)[\/latex]<\/p>\r\nFind the GCF of the first pair of terms.\r\n<p style=\"text-align: center;\">[latex]a^2=a\\cdot a[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]3a = 3\\cdot a[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]GCF= a[\/latex]<\/p>\r\nFactor the GCF, [latex]a[\/latex], out of the first group.\r\n<p style=\"text-align: center;\">[latex]a(a+3)+(5a+15)[\/latex]<\/p>\r\nFind the GCF of the second pair of terms.\r\n<p style=\"text-align: center;\">[latex]5a=5 \\cdot a[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]15= 5 \\cdot 3[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]GCF = 5[\/latex]<\/p>\r\nFactor [latex]5[\/latex] out of the second group.\r\n<p style=\"text-align: center;\">[latex]a(a+3)+5(a+3)[\/latex]<\/p>\r\nFactor out the common factor [latex](a+3)[\/latex] from the two terms. The [latex]a[\/latex] and the [latex]5[\/latex] become a binomial sum along with the common factor of [latex](a+3)[\/latex].\r\n<p style=\"text-align: center;\">[latex](a+3)(a+5)[\/latex]<\/p>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that with this process, we have been able to factor a four-term polynomial into the product of two binomials.\r\n\r\nThis process is called the <i>grouping technique<\/i> . Broken down into individual steps, here\u2019s how we apply this technique to a four-term polynomial (you can also follow this process in the example below).\r\n<div class=\"textbox\">\r\n<h3>How To: Given a 4-term polynomial expression, factor by grouping<\/h3>\r\n<ol>\r\n \t<li>Group the terms with common factors. Sometimes it will be necessary to rearrange the terms.<\/li>\r\n \t<li>Find the greatest common factor of each pair and factor it out.<\/li>\r\n \t<li>Look for the common binomial\u00a0between the factored terms.<\/li>\r\n \t<li>If the two terms share a common binomial factor, factor the common binomial\u00a0out of the groups.<\/li>\r\n<\/ol>\r\n<\/div>\r\nIt is worth noting that grouping can be used for polynomials with more than four terms, but the steps above and our examples will focus on four-term polynomials.\u00a0 Let\u2019s try factoring a few more. Note how there is a now a coefficient in front of the [latex]x^2[\/latex]\u00a0term. We will just consider this another factor when we are finding the GCF.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]2x^2+4x+5x+10[\/latex]\r\n\r\n[reveal-answer q=\"491917\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"491917\"]\r\n\r\nGroup terms of the polynomial into pairs.\r\n<p style=\"text-align: center;\">[latex](2x^2+4x)+(5x+10)[\/latex]<\/p>\r\nFactor out the like factor, [latex]2x[\/latex], from the first group, and factor out [latex]5[\/latex] from the second group.\r\n<p style=\"text-align: center;\">[latex]2x(x+2)+5(x+2)[\/latex]<\/p>\r\nLook for common factors between the factored forms of the paired terms. Here, the common factor is [latex](x+2)[\/latex]. Factor out the common factor, [latex](x+2)[\/latex], from both terms. Leaving [latex](2x+5)[\/latex] as the other factor.\r\n<p style=\"text-align: center;\">[latex](x+2)(2x+5)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video provides another example of factor by grouping.\r\n\r\n<iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/RR5nj7RFSiU?si=GSh05M4ZbK5Mvvis\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]3x^2 +3x -2x -2[\/latex]\r\n\r\n[reveal-answer q=\"171835\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"171835\"]\r\n\r\nGroup terms into pairs.\r\n<p style=\"text-align: center;\">[latex](3x^2+3x)+(-2x-2)[\/latex]<\/p>\r\nFactor the common factor of [latex]3x[\/latex] out of the first group and [latex]-2[\/latex] out of the second group. Notice what happens when a [latex]-2[\/latex], not a [latex]2[\/latex], is factored out.\r\n<p style=\"text-align: center;\">[latex]3x(x+1) -2(x+1)[\/latex]<\/p>\r\nFactor out the common factor, [latex](x+1)[\/latex], from both terms, leaving [latex](3x-2)[\/latex] as the other factor.\r\n<p style=\"text-align: center;\">[latex](x+1)(3x-2)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we present another example of factor by grouping.\r\n\r\n<iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0dvGmDGVC5U?si=s6Oo5a6IHoXoJVw9\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nFactor completely. [latex]4x^3+12x^2+x+3[\/latex]\r\n\r\n[reveal-answer q=\"197956\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"197956\"]\r\n\r\nGroup the expression into pairs.\r\n<p style=\"text-align: center;\">[latex](4x^3+12x^2) + (x+3)[\/latex]<\/p>\r\nThe GCF of the first group is [latex]4x^2[\/latex] so let's factor [latex]4x^2[\/latex] out of the first group. What is the GCF of the second group? The common factor of the second pair is 1. There is no other factors that the terms have in common. We can write that as\r\n<p style=\"text-align: center;\">[latex]4x^2(x+3)+ 1(x+3)[\/latex]<\/p>\r\nWriting the [latex]1[\/latex] in front, makes the factoring more apparent. Factor out the common factor, [latex](x+3)[\/latex], from both terms, leaving [latex](4x^2+1)[\/latex] as the other factor.\r\n<p style=\"text-align: center;\">[latex](x+3)(4x^2 +1)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 id=\"Trinomial1\">Factoring Trinomials with Leading Coefficient of 1<\/h2>\r\nTrinomials are polynomials with three terms. We are going to show you a method for factoring a trinomial whose leading coefficient is\u00a0[latex]1[\/latex]. \u00a0Although we should always begin by looking for a GCF, factoring out the GCF is not the only way that trinomials\u00a0can be factored. The trinomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of\u00a0[latex]1[\/latex], but it can be written as the product of the binomial factors.\r\n\r\nThe last example in the previous section asked us to graph [latex]g(x)=x^2+5x+6[\/latex]. It appeared that it would be easier to find the [latex]x[\/latex]-intercept(s), vertex, and axis of symmetry if we could change the function to intercept form. How do we change this function from general form [latex]f(x)=ax^2+bx+c[\/latex] to intercept (factored) form [latex]f(x)=(x-p)(x-q)[\/latex]? We need to learn how to factor trinomials.\r\n<p style=\"text-align: center;\">We will start by factoring the polynomial [latex]1x^2+5x+6[\/latex].<\/p>\r\nLet's identify the values for [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex]. Looking at the given polynomial, [latex]a=1[\/latex], [latex]b=5[\/latex], and [latex]c=6[\/latex].\r\n\r\nWe need to find two integers [latex]m[\/latex] and [latex]n[\/latex] whose product is [latex]c[\/latex] and whose sum is [latex]b[\/latex]. In other words, let's think of two integers that multiply to be [latex]6[\/latex] and add to be [latex]5[\/latex]. The factors of 6 are\r\n<table style=\"width: 30%; font-size: 110%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, 6; -1, -6; -2, -3; and 2, 3. The entries in the second column are: 7, -7, -5, and 5.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]6[\/latex]<\/th>\r\n<th>Sum of 5<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,6[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,-6[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,-3[\/latex]<\/td>\r\n<td>[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><span style=\"background-color: #ffff00;\">[latex]2,3[\/latex]<\/span><\/td>\r\n<td><span style=\"background-color: #ffff00;\">[latex]5[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhich pair of factors add to be [latex]5[\/latex]? The only way to get a sum of 5 is by adding the [latex]2[\/latex] and [latex]3[\/latex] ([latex]2+3=5[\/latex]). We have identified [latex]m[\/latex] as [latex]2[\/latex] and [latex]n[\/latex] as [latex]3[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} ax^2+bx+c &amp;= (x+m)(x+n)\\\\ &amp;= (x+2)(x+3) \\end{align}[\/latex]<\/p>\r\nNOTE: The answer could also be written as [latex](x+3)(x+2)[\/latex]. We can change the order of the factors since multiplication is commutative.\r\n\r\nTo check if it is factored correctly, multiply the two binomials:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\left(x+2\\right)\\left(x+3\\right) &amp;= x^2+3x+2x+6\\\\ &amp;= x^2+5x+6 \\end{align}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nIn general, for a trinomial of the form [latex]{x}^{2}+bx+c[\/latex], you can factor a trinomial with leading coefficient\u00a0[latex]1[\/latex] by finding two integers, [latex]m[\/latex] and [latex]n[\/latex], whose product is [latex]c[\/latex] and whose sum is [latex]b[\/latex].\r\n\r\n<\/div>\r\nLet us put this idea to practice with the following example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]{x}^{2}+2x - 15[\/latex]\r\n[reveal-answer q=\"44696\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44696\"]\r\n\r\nWe have a trinomial with leading coefficient [latex]1, b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two integers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table style=\"width: 20%; font-size: 110%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]-15[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,-15[\/latex]<\/td>\r\n<td>[latex]-14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,15[\/latex]<\/td>\r\n<td>[latex]14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-5[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><span style=\"background-color: #ffff00;\">[latex]-3,5[\/latex]<\/span><\/td>\r\n<td><span style=\"background-color: #ffff00;\">[latex]2[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow that we have identified [latex]m[\/latex] and [latex]n[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we present two more examples of factoring a trinomial with a leading coefficient of [latex]1[\/latex].\r\n\r\nhttps:\/\/youtu.be\/-SVBVVYVNTM\r\n\r\nTo summarize our process, consider the following steps:\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]c[\/latex].<\/li>\r\n \t<li>Find [latex]m[\/latex] and [latex]n[\/latex], a pair of factors with a product of [latex]c[\/latex] and with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Write the factored expression as [latex]\\left(x+m\\right)\\left(x+n\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn our next example, we show that when [latex]c[\/latex] is negative, either [latex]m[\/latex] or [latex]n[\/latex] will be negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]x^2+x-12[\/latex]\r\n\r\n[reveal-answer q=\"205737\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"205737\"]\r\n\r\nConsider all the combinations of integers whose product is\u00a0[latex]-12[\/latex] and list their sum.\r\n<table style=\"width: 30%; height: 84px; font-size: 110%;\">\r\n<thead>\r\n<tr style=\"height: 12px;\">\r\n<th style=\"height: 12px;\">Factors of [latex]\u221212[\/latex]<\/th>\r\n<th style=\"height: 12px;\">Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]1, \u221212[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]2, \u22126[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]\u22124[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]3, \u22124[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]\u22121[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\"><span style=\"background-color: #ffff00;\">[latex]4, \u22123[\/latex]<\/span><\/td>\r\n<td style=\"height: 12px;\"><span style=\"background-color: #ffff00;\">[latex]1[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]6, \u22122[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]12, \u22121[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]11[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nChoose the values whose sum is\u00a0[latex]+1[\/latex]:\u00a0\u00a0[latex]m=4[\/latex] and [latex]n=\u22123[\/latex], and place them into a product of binomials.\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]{x}^{2}-7x+6[\/latex]\r\n[reveal-answer q=\"662468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"662468\"]\r\n\r\nList the factors of\u00a0[latex]6[\/latex]. Note that the b term is negative, so we will need to consider negative numbers in our list.\r\n<table style=\"width: 20%; font-size: 110%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]6[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,6[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2, 3[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><span style=\"background-color: #ffff00;\">[latex]-1, -6[\/latex]<\/span><\/td>\r\n<td><span style=\"background-color: #ffff00;\">[latex]-7[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2, -3[\/latex]<\/td>\r\n<td>[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nChoose the pair that sum to [latex]-7[\/latex]: \u00a0[latex]m=-1[\/latex], [latex]n=-6[\/latex]\r\n\r\nWrite the pair as constant terms in a product of binomials.\r\n\r\n[latex]\\left(x-1\\right)\\left(x-6\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the last example, the [latex]b[\/latex] term was negative and the [latex]c[\/latex] term was positive. This will always mean that if it can be factored, [latex]m[\/latex] and [latex]n[\/latex] will both be negative.\r\n<h2 id=\"Trinomialnot1\">Factoring trinomials with Leading Coefficient not 1<\/h2>\r\nTrinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by separating the [latex]x[\/latex] term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be factored as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as a 4-term polynomial [latex]2{x}^{2}+2x+3x+3[\/latex]. The process to do this is explained in the next example. We will \u00a0use factor by grouping to complete the factorization of the 4-term polynomial resulting in [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.\r\n<div class=\"textbox\">\r\n<h3>AC-method of Factoring<\/h3>\r\nTo factor a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] by using the AC-Method, we find two integers with a product of [latex]a\\cdot c[\/latex] and a sum of [latex]b[\/latex]. We use these integers to separate the [latex]x[\/latex] term into the sum of two terms (making it a 4-term polynomial). Then factor the polynomial by grouping.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.\r\n\r\n[reveal-answer q=\"806328\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"806328\"]\r\n\r\nWe have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]a\\cdot c=-30[\/latex]. We need to find two integers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table style=\"width: 20%; height: 84px; font-size: 110%;\" summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\r\n<thead>\r\n<tr style=\"height: 12px;\">\r\n<th style=\"height: 12px;\">Factors of [latex]-30[\/latex]<\/th>\r\n<th style=\"height: 12px;\">Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]1,-30[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]-29[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]-1,30[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]29[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]2,-15[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]-13[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]-2,15[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]13[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\">[latex]3,-10[\/latex]<\/td>\r\n<td style=\"height: 12px;\">[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"height: 12px;\"><span style=\"background-color: #ffff00;\">[latex]-3,10[\/latex]<\/span><\/td>\r\n<td style=\"height: 12px;\"><span style=\"background-color: #ffff00;\">[latex]7[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]m=-3[\/latex] and [latex]n=10[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align}5{x}^{2}-3x+10x-6 &amp;&amp; \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right) &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\\r\n\\left(5x - 3\\right)\\left(x+2\\right) &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>\r\n\r\nWe can check that the factorization is correct by multiplying the two binomials together.\r\n\r\n<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: AC-Method - Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex],<\/h3>\r\n<ol>\r\n \t<li>Factor out the GCF from all terms, if there is a GCF.<\/li>\r\n \t<li>List factors of [latex]a\\cdot c[\/latex].<\/li>\r\n \t<li>Find [latex]m[\/latex] and [latex]n[\/latex], a pair of factors of [latex]a\\cdot c[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Rewrite the original expression as a 4-term polynomial.<\/li>\r\n \t<li>Factor out the GCF of first two terms.<\/li>\r\n \t<li>Factor out the GCF of last two terms.<\/li>\r\n \t<li>Factor out the GCF of the expression.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]6x^2+11x+4[\/latex]\r\n\r\n[reveal-answer q=\"864920\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"864920\"]\r\n\r\nWe have a trinomial with [latex]a=6,b=11[\/latex], and [latex]c=4[\/latex]. First, determine [latex]a\\cdot c=24[\/latex]. We need to find two numbers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]24[\/latex] and a sum of [latex]11[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table style=\"width: 20%; font-size: 110%;\" summary=\"A table with 9 rows and 2 columns. The first column is labeled: Factors of 24 while the second is labeled: Sum of Factors. The entries in the first column are: 1, 24; -1, -24; 2, 12; -2, -12; 3, 8; -3, -8; 4, 6; -4, -6. The entries in the second column are: 25, -25, 14, -14, 11, -11, 10 and -10.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]24[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,24[\/latex]<\/td>\r\n<td>[latex]25[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,-24[\/latex]<\/td>\r\n<td>[latex]-25[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2,12[\/latex]<\/td>\r\n<td>[latex]14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,-12[\/latex]<\/td>\r\n<td>[latex]-14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><span style=\"background-color: #ffff00;\">[latex]3,8[\/latex]<\/span><\/td>\r\n<td><span style=\"background-color: #ffff00;\">[latex]11[\/latex]<\/span><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,-8[\/latex]<\/td>\r\n<td>[latex]-11[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4,6[\/latex]<\/td>\r\n<td>[latex]10[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-4,-6[\/latex]<\/td>\r\n<td>[latex]-10[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]m=3[\/latex] and [latex]n=8[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align}6{x}^{2}+3x+8x+4 &amp;&amp; \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\\r\n3x\\left(2x +1\\right)+4\\left(2x +1\\right) &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\ \\left(2x +1\\right)\\left(3x+4\\right) &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>\r\n\r\nWe can check that the factorization is correct by multiplying the two binomials together.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nFactor completely. [latex]10x^2-7x-6[\/latex]\r\n\r\n[reveal-answer q=\"559688\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"559688\"]\r\n\r\nWe have a trinomial with [latex]a=10,b=-7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]a\\cdot c=-60[\/latex]. We need to find two integers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]-60[\/latex] and a sum of [latex]-7[\/latex].\r\n\r\nYou can begin by listing out all pairs that multiply to [latex]-60[\/latex] if you find this helpful.\r\n\r\nSo [latex]m=-12[\/latex] and [latex]n=5[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align}10{x}^{2}-12x+5x-6 &amp;&amp; \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\\r\n2x\\left(5x-6\\right)+1\\left(5x-6\\right) &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\ \\left(5x-6\\right)\\left(2x+1\\right) &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>\r\n\r\nWe can check that the factorization is correct by multiplying the two binomials together.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]5x^2-x+2[\/latex]\r\n\r\nWe have a trinomial with [latex]a=5,b=-1[\/latex], and [latex]c=-2[\/latex]. First, determine [latex]a\\cdot c=10[\/latex]. We need to find two integers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]10[\/latex] and a sum of [latex]-1[\/latex].\r\n<table style=\"width: 20%; font-size: 110%;\" summary=\"A table with 9 rows and 2 columns. The first column is labeled: Factors of 24 while the second is labeled: Sum of Factors. The entries in the first column are: 1, 24; -1, -24; 2, 12; -2, -12; 3, 8; -3, -8; 4, 6; -4, -6. The entries in the second column are: 25, -25, 14, -14, 11, -11, 10 and -10.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]10[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,10[\/latex]<\/td>\r\n<td>[latex]11[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,-10[\/latex]<\/td>\r\n<td>[latex]-11[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2,5[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,-5[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe have exhausted all the possible integers that multiply to be [latex]10[\/latex] and none of the combinations of factors will add to be [latex]-1[\/latex]. Therefore, the polynomial is not factorable and is said to be <strong>prime<\/strong>.\r\n\r\nAnswer: Not Factorable or Prime\r\n\r\n<\/div>\r\nThe following video has another example about factoring trinomials when the leading coefficient is not 1.\r\n\r\n<iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/agDaQ_cZnNc?si=SqrqME5CdbK4p6O9\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\nIn some situations, [latex]a[\/latex] is negative, as in [latex]-4h^2+11h+3[\/latex]. It often makes sense to factor out [latex]-1[\/latex] as the first step, making the polynomial easier to factor.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]-4h^2+11h+3[\/latex]\r\n\r\n[reveal-answer q=\"745970\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"745970\"]\r\n\r\nFactor [latex]-1[\/latex] out of the trinomial. Notice that the signs of all three terms have changed.\r\n<p style=\"text-align: center;\">[latex]-1(4h^2-11h-3)[\/latex]<\/p>\r\nWe now have a trinomial with [latex]a=4,b=-11[\/latex], and [latex]c=-3[\/latex]. First, determine [latex]a\\cdot c=-12[\/latex]. We need to find two numbers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]-12[\/latex] and a sum of [latex]-11[\/latex].\r\n\r\nYou can begin by listing out all pairs that multiply to [latex]-12[\/latex] if you find this helpful.\r\n\r\nSo [latex]m=-12[\/latex] and [latex]n=1[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align}-1(4{h}^{2}-12h+1h-3) &amp;&amp; \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\\r\n-1[\\,4h\\left(h-3\\right)+1\\left(h-3\\right)]\\, &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\ -1\\left(h-3\\right)\\left(4h+1\\right) &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>\r\n\r\nWe can check that the factorization is correct by multiplying the two binomials together.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]16x^4-12x^3-10x^2[\/latex]\r\n\r\n[reveal-answer q=\"945393\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"945393\"]\r\n\r\nIs there a GCF that all three terms have in common? Yes, [latex]2x^2[\/latex] is a common factor of all three. We will factor this out first.\r\n<p style=\"text-align: center;\">[latex]2x^2(8x^2-6x-5)[\/latex]<\/p>\r\nThe trinomial that we need to factor has coefficients of [latex]a=8,b=-6[\/latex], and [latex]c=-5[\/latex]. First, determine [latex]a\\cdot c=-40[\/latex]. We need to find two integers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]-40[\/latex] and a sum of [latex]-6[\/latex].\r\n\r\nYou can begin by listing out all pairs that multiply to [latex]-40[\/latex] if you find this helpful.\r\n\r\nSo [latex]m=-10[\/latex] and [latex]n=4[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align} 2x^2(8x^2-10x+4x-5) &amp;&amp; \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\\r\n2x^2[\\,2x\\left(4x-5\\right)+1\\left(4x-5\\right)]\\, &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\ 2x^2\\left(4x-5\\right)\\left(2x+1\\right) &amp;&amp; \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>\r\n\r\nWe can check that the factorization is correct by multiplying the two binomials together.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2 id=\"special\">Factoring Special Cases<\/h2>\r\n<h3>Factoring a Perfect Square Trinomial<\/h3>\r\nWhat is a perfect square? A perfect square is the product of an integer multiplied by itself. For example, [latex]4 \\cdot 4 = 16[\/latex] So, [latex]16[\/latex] is a perfect square. Other examples of perfect squares are, [latex]1, 4, 9, 16, 25, 36, 49, 64, 81[\/latex], and [latex]100[\/latex]. A perfect square trinomial is a trinomial that can be written in factored form as a binomial squared. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}{\\left(a+b\\right)}^{2}\\hfill &amp; =&amp; {a}^{2}+2ab+{b}^{2}\\hfill \\\\ &amp; \\text{and}&amp; \\\\ \\hfill {\\left(a-b\\right)}^{2}&amp; =&amp; {a}^{2}-2ab+{b}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div style=\"text-align: left;\">We can use these same equations to factor any perfect square trinomial.<\/div>\r\n<div class=\"textbox\">\r\n<h3>Perfect Square Trinomials<\/h3>\r\nA perfect square trinomial can be written, in factored form, as a binomial squared:\r\n<div style=\"text-align: center;\">[latex]\\large{{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex]\\large{{a}^{2}-2ab+{b}^{2}={\\left(a-b\\right)}^{2}}[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]25{x}^{2}+20x+4[\/latex]\r\n\r\n[reveal-answer q=\"114092\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"114092\"]\r\n\r\nFirst, notice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex].\r\n\r\nIn using the equations above to help you factor this, we can say that [latex]a=5x[\/latex] and [latex]b=2[\/latex].\r\n\r\nNext, check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex].\r\n\r\nTherefore, the trinomial is a perfect square trinomial and can be written as [latex]{(a+b)}^2={\\left(5x+2\\right)}^{2}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a perfect square trinomial, factor it into the square of a binomial<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Confirm that the middle term is twice the product of [latex]a\\cdot b[\/latex].<\/li>\r\n \t<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex] or [latex]{(a-b)}^{2}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn the following video, we provide another short description of perfect square trinomials and shows how to factor them using a formula.\r\n\r\n<iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UMCVGDTxxTI?si=yqGYn8z1CD59WvPn\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<h3>Factoring a Difference of Squares<\/h3>\r\nA difference of squares can be factored into two binomials but has only two terms. \u00a0Let\u2019s start from the product of two special binomials to see the pattern.\r\n\r\nConsider the product of the following two binomials, which are identical except for the operation: [latex](x-2)(x+2)[\/latex]. If we multiply them together, the middle term zeros out.\r\n<p style=\"text-align: center;\">Multiply. [latex](x-2)(x+2)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;= x^2-2x+2x-2^2\\\\ &amp;= x^2-2^2\\\\ &amp;= x^2-4 \\end{align}[\/latex]<\/p>\r\nThe polynomial [latex]x^2-4[\/latex] is called a difference of squares because each term is a perfect square and the operation connecting them is subtraction. A difference of squares will always factor in the following way:\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n[latex]\\\\[\/latex]\r\n\r\nWe can use this equation to factor any differences of squares.\r\n<div class=\"textbox\">\r\n<h3>Differences of Squares<\/h3>\r\nA difference of squares can be rewritten as two factors containing the same terms but opposite signs.\r\n<div style=\"text-align: center;\">[latex]\\large{{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)}[\/latex]<\/div>\r\n<\/div>\r\n<div>The following video shows examples of factoring a difference of squares.<\/div>\r\n<iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Li9IBp5HrFA?si=sPSn9f6DDEFLsmlw\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<div><\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a difference of squares, factor it into binomials<\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Determine [latex]a[\/latex] and [latex]b[\/latex].<\/li>\r\n \t<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely. [latex]9{x}^{2}-25[\/latex]\r\n\r\n[reveal-answer q=\"830417\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"830417\"]\r\n\r\nNotice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex].\r\n\r\nThis means that [latex]a=3x[\/latex] and [latex]b=5[\/latex].\r\n\r\nThe polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].\r\n\r\nYou can <strong>check<\/strong> that it is factored correctly by multiplying the two binomials.\r\n\r\n[latex]\\begin{align}(3x+5)(3x-5)\\\\ &amp;= 9x^2-15x+15x-5^2\\\\ &amp;= 9x^2-5^2\\\\ &amp;= 9x^2-25 \\end{align}[\/latex]\r\n\r\nMultiplying the binomials results in the polynomial that we started with so, it is indeed factored correctly. Our factored answer is:\r\n<p style=\"text-align: center;\">[latex](3x+5)(3x-5)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nFactor completely. [latex]18x^2-2x[\/latex]\r\n\r\n[reveal-answer q=\"633557\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"633557\"]\r\n\r\nNotice that the terms of this binomial are not perfect squares. However, they do share a common factor of [latex]2x[\/latex] which can be factored out.\r\n<p style=\"text-align: center;\">[latex]2x(9x^2-1)[\/latex]<\/p>\r\nThe resulting binomial is now a difference of squares. Since [latex]9x^2=(3x)^2[\/latex] and [latex]1=1^2[\/latex], we have [latex]a=3x[\/latex] and [latex]b=1[\/latex]. Applying our difference of squares equation, and remembering the fact of [latex]2x[\/latex] in front, this factors as\r\n<p style=\"text-align: center;\">[latex]2x(3x+1)(3x-1)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Summary<\/h2>\r\nIn all factoring problems, we first look for a common factor that can be factored out of all terms. After achieving this (if possible), we have now learned special factoring techniques in this section.\r\n\r\nIf a trinomial is of the form of a perfect square trinomial, it factors as\r\n<div style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">or<\/div>\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-2ab+{b}^{2}={\\left(a-b\\right)}^{2}[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div>If a binomial is of the form of a difference of squares, [latex]a^2-b^2[\/latex], it will factor as<\/div>\r\n<div style=\"text-align: center;\">[latex]a^2-b^2= (a+b)(a-b)[\/latex]<\/div>\r\n<div><\/div>\r\n<div>If a binomial is of the form of a sum of squares, [latex]a^2+b^2[\/latex], it will not factor and is considered a prime polynomial.<\/div>\r\n<div><\/div>\r\n<div><\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Review Objectives<\/h3>\n<ul>\n<li>Factor out the greatest common factor of a polynomial<\/li>\n<li>Factor by grouping<\/li>\n<li>Factor trinomials (when leading coefficient is one and when leading coefficient is not one)<\/li>\n<li>Factor perfect square trinomials and difference of squares<\/li>\n<\/ul>\n<\/div>\n<p>Click on the following links to move to different types of factoring on this page.<br \/>\n<a class=\"anchorLink\" href=\"#gcf\">Factor out the GCF<\/a><br \/>\n<a class=\"anchorLink\" href=\"#Grouping\">Factoring 4-term Polynomials by Grouping<\/a><br \/>\n<a class=\"anchorLink\" href=\"#Trinomial1\">Factoring Trinomials with Leading Coefficient of 1<\/a><br \/>\n<a class=\"anchorLink\" href=\"#Trinomialnot1\">Factoring trinomials with Leading Coefficient not 1<\/a><br \/>\n<a class=\"anchorLink\" href=\"#special\">Factoring Special Cases<\/a><\/p>\n<h2>Factoring Basics<\/h2>\n<p><strong>Factors<\/strong> are the building blocks of multiplication. They are the integers that you can multiply together to produce another integer: [latex]2[\/latex] and\u00a0[latex]10[\/latex] are factors of\u00a0[latex]20[\/latex], as are\u00a0[latex]4, 5, 1, 20[\/latex]. To factor an integer is to rewrite it as a product. [latex]20=4\\cdot{5}[\/latex] or [latex]20=1\\cdot{20}[\/latex]. In algebra, we use the word factor as both a noun \u2013 something being multiplied \u2013 and as a verb \u2013 the action of rewriting a sum or difference as a product.\u00a0<strong>Factoring<\/strong> is very helpful in simplifying expressions and solving equations involving\u00a0polynomials.<\/p>\n<p>The <strong>greatest common factor<\/strong> (GCF) of two integers is the largest common factor to each integer. For instance, [latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest integer that divides evenly into both [latex]16[\/latex] and [latex]20[\/latex]. The GCF of polynomials works the same way: [latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest polynomial that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].<\/p>\n<p><strong>When factoring a polynomial expression, our first step should be to check for a GCF.<\/strong> Look for the GCF of the coefficients, and then look for the GCF of the variables.<\/p>\n<div class=\"textbox\">\n<h3>Greatest Common Factor<\/h3>\n<p>The <strong>greatest common factor<\/strong> (GCF) of a group of given polynomials is\u00a0the largest integer and highest degree of each variable that will divide evenly into each term of the polynomial.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Find the greatest common factor of [latex]25b^{3}[\/latex] and [latex]10b^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q210634\">Show Solution<\/span><\/p>\n<div id=\"q210634\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,25b^{3}=5\\cdot5\\cdot{b}\\cdot{b}\\cdot{b}\\\\\\,\\,10b^{2}=5\\cdot2\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5b^{2}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the example above, the monomials have the factors\u00a0[latex]5[\/latex], [latex]b[\/latex], and [latex]b[\/latex] in common, which means their greatest common factor is [latex]5\\cdot{b}\\cdot{b}[\/latex], or simply [latex]5b^{2}[\/latex].<\/p>\n<p>The video that follows gives an example of finding the greatest common factor of two monomials with only one variable.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Determine the GCF of Two Monomials (One Variables)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/EhkVBXRBC2s?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Sometimes you may encounter a polynomial with more than one variable, so it is important to check whether both variables are part of the GCF. In the next example, we find the GCF of two terms which both contain two variables.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Find the greatest common factor of [latex]81c^{3}d[\/latex] and [latex]45c^{2}d^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q930504\">Show Solution<\/span><\/p>\n<div id=\"q930504\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,81c^{3}d=3\\cdot3\\cdot3\\cdot3\\cdot{c}\\cdot{c}\\cdot{c}\\cdot{d}\\\\45c^{2}d^{2}=3\\cdot3\\cdot5\\cdot{c}\\cdot{c}\\cdot{d}\\cdot{d}\\\\\\,\\,\\,\\,\\text{GCF}=3\\cdot3\\cdot{c}\\cdot{c}\\cdot{d}\\\\\\,\\,\\,\\,\\text{GCF}=9c^{2}d\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The video that follows shows another example of finding the greatest common factor of two monomials with more than one variable.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex: Determine the GCF of Two Monomials (Two Variables)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/GfJvoIO3gKQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You might be picking up on a useful shortcut for the variables. Notice that (assuming the variable is in common) we always select the smaller power of the variable.<\/p>\n<h2 id=\"gcf\">Factoring out the Greatest Common Factor (GCF)<\/h2>\n<p>Now that you have practiced identifying the GCF of terms with one and two variables, we can apply this idea to factoring\u00a0the GCF out of a polynomial. Notice that\u00a0the instructions are now &#8220;Factor&#8221; instead of &#8220;Find the greatest common factor.&#8221;<\/p>\n<p>To factor a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property to rewrite the polynomial in factored form.<\/p>\n<div class=\"textbox shaded\">\n<h3>Distributive Property Forward and Backward<\/h3>\n<p>Forward: Product of a number and a sum: [latex]a\\left(b+c\\right)=a\\cdot{b}+a\\cdot{c}[\/latex]. You can say that \u201c[latex]a[\/latex] is being distributed over [latex]b+c[\/latex].\u201d<\/p>\n<p>Backward: Sum of the products: [latex]a\\cdot{b}+a\\cdot{c}=a\\left(b+c\\right)[\/latex]. Here you can say that \u201c[latex]a[\/latex] is being factored out.\u201d<\/p>\n<p>We first learned that we could distribute a factor over a sum or difference, now we are learning that we can &#8220;undo&#8221; the distributive property with factoring.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]25b^{3}+10b^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q716902\">Show Solution<\/span><\/p>\n<div id=\"q716902\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the GCF. From a previous example, you found the GCF of [latex]25b^{3}[\/latex] and [latex]10b^{2}[\/latex] to be [latex]5b^{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,25b^{3}=5\\cdot5\\cdot{b}\\cdot{b}\\cdot{b}\\\\\\,\\,10b^{2}=5\\cdot2\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5\\cdot{b}\\cdot{b}=5b^{2}\\end{array}[\/latex]<\/p>\n<p>Rewrite each term with the GCF as one factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3} = 5b^{2}\\cdot5b\\\\10b^{2}=5b^{2}\\cdot2\\end{array}[\/latex]<\/p>\n<p>Rewrite the polynomial using the factored terms in place of the original terms.<\/p>\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b\\right)+5b^{2}\\left(2\\right)[\/latex]<\/p>\n<p>Factor out the [latex]5b^{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b+2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The factored form of the polynomial [latex]25b^{3}+10b^{2}[\/latex] is [latex]5b^{2}\\left(5b+2\\right)[\/latex]. You can check this by doing the multiplication. [latex]5b^{2}\\left(5b+2\\right)=25b^{3}+10b^{2}[\/latex].<\/p>\n<p>Note that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.<\/p>\n<p>For example:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3}+10b^{2}=5\\left(5b^{3}+2b^{2}\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }5\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5b^{2}\\left(5b+2\\right) \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }b^{2}\\end{array}[\/latex]<\/p>\n<p>Notice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.<\/p>\n<p>In the following video, we show two more examples of how to find and factor the GCF from binomials.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 1:  Identify GCF and Factor a Binomial\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/25_f_mVab_4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox\">\n<h3>How To: Given a polynomial expression, factor out the greatest common factor<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Identify the GCF of the coefficients.<\/li>\n<li>Identify the GCF of the variables.<\/li>\n<li>Write together to find the GCF of the expression.<\/li>\n<li>Determine what the GCF needs to be multiplied by to obtain each term in the expression.<\/li>\n<li>Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.<\/li>\n<\/ol>\n<\/div>\n<p>We will show a few examples of finding the GCF of a polynomial with several terms and sometimes with two variables. No matter how many terms the polynomial has, you can use the same technique described above to factor out its GCF.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor out the GCF. [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q112050\">Show Solution<\/span><\/p>\n<div id=\"q112050\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find the GCF of the expression.<\/p>\n<p>The GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. The GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions in the form [latex]{x}^{n}[\/latex] will always be the variable with the exponent of lowest degree.) And the GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. Put these together to find the GCF of the polynomial, [latex]3xy[\/latex].<\/p>\n<p>Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that [latex]3xy\\left(2{x}^{2}{y}^{2}\\right)=6{x}^{3}{y}^{3},3xy\\left(15xy\\right)=45{x}^{2}{y}^{2}[\/latex], and [latex]3xy\\left(7\\right)=21xy[\/latex].<\/p>\n<p>Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by:<\/p>\n<p>[latex]3xy\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]<\/p>\n<p>After factoring, we can check our work by multiplying. Use the distributive property to check.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will see two more\u00a0example of how to find and factor out the greatest common factor of a polynomial.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 2:  Identify GCF and Factor a Trinomial\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/3f1RFTIw2Ng?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next example, the leading coefficient is negative. In this case, it is convention to include the negative with the GCF. In addition, we will see that this can assist us in other ways as well.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor out the GCF. [latex]-24x^8+32x^5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q707771\">Show Solution<\/span><\/p>\n<div id=\"q707771\" class=\"hidden-answer\" style=\"display: none\">\n<p>The GCF of [latex]24[\/latex] and [latex]32[\/latex] is 8, while the GCF of [latex]x^8[\/latex] and [latex]x^5[\/latex] is [latex]x^5[\/latex]. However,\u00a0because the leading coefficient is negative, we will include the negative in the overall GCF.<\/p>\n<p style=\"text-align: center;\">[latex]GCF = -8x^5[\/latex]<\/p>\n<p>Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that<\/p>\n<p style=\"text-align: center;\">[latex]-24x^8=-8x^5(3x^3)[\/latex] and [latex]32x^5=-8x^5(-4)[\/latex]<\/p>\n<p>Thus, factoring out the GCF gives us,<\/p>\n<p style=\"text-align: center;\">[latex]-8x^5(3{x}^{3}-4)[\/latex]<\/p>\n<p>Note that the effect of factoring out a negative term is that all signs of the remaining polynomial will be the opposite of the original terms.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Another interesting situation we may encounter is where the GCF could even include an entire expression containing multiple terms, as shown in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor out the GCF. [latex]3x^2(2x-5)+7(2x-5)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q497376\">Show Solution<\/span><\/p>\n<div id=\"q497376\" class=\"hidden-answer\" style=\"display: none\">\n<p>Resisting the temptation to distribute, we can view this as a polynomial with two terms: [latex]3x^2(2x-5)[\/latex] and [latex]7(2x-5)[\/latex]. The GCF of [latex]3[\/latex] and [latex]7[\/latex] is simply [latex]1[\/latex] and the terms do not share any power of [latex]x[\/latex] in common. However, each term contains the expression [latex](2x-5)[\/latex], so this will be the GCF.<\/p>\n<p style=\"text-align: center;\">[latex]GCF=(2x-5)[\/latex]<\/p>\n<p>We can factor this out using the same strategy we used on all previous problems, writing the expression as a product of the GCF with a sum comprised of what we would need to multiply by to obtain each original term. So, factoring out the GCF yields.<\/p>\n<p style=\"text-align: center;\">[latex](2x-5)(3{x}^{2}+7)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2 id=\"Grouping\">Factor by Grouping<\/h2>\n<p>Factoring by grouping is a technique that allows us to factor a polynomial whose terms don\u2019t all share a GCF. In the following example, we will introduce you to the technique. Remember, some of the main reasons we want to be able to factor well \u00a0is because it will help solve polynomial equations and graph quadratic functions more easily.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]a^2+3a+5a+15[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q835783\">Show Solution<\/span><\/p>\n<div id=\"q835783\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that there is no GCF other than 1 for all four terms. Resist the temptation to combine the like terms, as instead we are going to group the problem into two pairs of terms and then search for the GCF of each.<\/p>\n<p style=\"text-align: center;\">[latex](a^2+3a)+(5a+15)[\/latex]<\/p>\n<p>Find the GCF of the first pair of terms.<\/p>\n<p style=\"text-align: center;\">[latex]a^2=a\\cdot a[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]3a = 3\\cdot a[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]GCF= a[\/latex]<\/p>\n<p>Factor the GCF, [latex]a[\/latex], out of the first group.<\/p>\n<p style=\"text-align: center;\">[latex]a(a+3)+(5a+15)[\/latex]<\/p>\n<p>Find the GCF of the second pair of terms.<\/p>\n<p style=\"text-align: center;\">[latex]5a=5 \\cdot a[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]15= 5 \\cdot 3[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]GCF = 5[\/latex]<\/p>\n<p>Factor [latex]5[\/latex] out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]a(a+3)+5(a+3)[\/latex]<\/p>\n<p>Factor out the common factor [latex](a+3)[\/latex] from the two terms. The [latex]a[\/latex] and the [latex]5[\/latex] become a binomial sum along with the common factor of [latex](a+3)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex](a+3)(a+5)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice that with this process, we have been able to factor a four-term polynomial into the product of two binomials.<\/p>\n<p>This process is called the <i>grouping technique<\/i> . Broken down into individual steps, here\u2019s how we apply this technique to a four-term polynomial (you can also follow this process in the example below).<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a 4-term polynomial expression, factor by grouping<\/h3>\n<ol>\n<li>Group the terms with common factors. Sometimes it will be necessary to rearrange the terms.<\/li>\n<li>Find the greatest common factor of each pair and factor it out.<\/li>\n<li>Look for the common binomial\u00a0between the factored terms.<\/li>\n<li>If the two terms share a common binomial factor, factor the common binomial\u00a0out of the groups.<\/li>\n<\/ol>\n<\/div>\n<p>It is worth noting that grouping can be used for polynomials with more than four terms, but the steps above and our examples will focus on four-term polynomials.\u00a0 Let\u2019s try factoring a few more. Note how there is a now a coefficient in front of the [latex]x^2[\/latex]\u00a0term. We will just consider this another factor when we are finding the GCF.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]2x^2+4x+5x+10[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q491917\">Show Solution<\/span><\/p>\n<div id=\"q491917\" class=\"hidden-answer\" style=\"display: none\">\n<p>Group terms of the polynomial into pairs.<\/p>\n<p style=\"text-align: center;\">[latex](2x^2+4x)+(5x+10)[\/latex]<\/p>\n<p>Factor out the like factor, [latex]2x[\/latex], from the first group, and factor out [latex]5[\/latex] from the second group.<\/p>\n<p style=\"text-align: center;\">[latex]2x(x+2)+5(x+2)[\/latex]<\/p>\n<p>Look for common factors between the factored forms of the paired terms. Here, the common factor is [latex](x+2)[\/latex]. Factor out the common factor, [latex](x+2)[\/latex], from both terms. Leaving [latex](2x+5)[\/latex] as the other factor.<\/p>\n<p style=\"text-align: center;\">[latex](x+2)(2x+5)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video provides another example of factor by grouping.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/RR5nj7RFSiU?si=GSh05M4ZbK5Mvvis\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]3x^2 +3x -2x -2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q171835\">Show Solution<\/span><\/p>\n<div id=\"q171835\" class=\"hidden-answer\" style=\"display: none\">\n<p>Group terms into pairs.<\/p>\n<p style=\"text-align: center;\">[latex](3x^2+3x)+(-2x-2)[\/latex]<\/p>\n<p>Factor the common factor of [latex]3x[\/latex] out of the first group and [latex]-2[\/latex] out of the second group. Notice what happens when a [latex]-2[\/latex], not a [latex]2[\/latex], is factored out.<\/p>\n<p style=\"text-align: center;\">[latex]3x(x+1) -2(x+1)[\/latex]<\/p>\n<p>Factor out the common factor, [latex](x+1)[\/latex], from both terms, leaving [latex](3x-2)[\/latex] as the other factor.<\/p>\n<p style=\"text-align: center;\">[latex](x+1)(3x-2)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we present another example of factor by grouping.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0dvGmDGVC5U?si=s6Oo5a6IHoXoJVw9\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Factor completely. [latex]4x^3+12x^2+x+3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q197956\">Show Solution<\/span><\/p>\n<div id=\"q197956\" class=\"hidden-answer\" style=\"display: none\">\n<p>Group the expression into pairs.<\/p>\n<p style=\"text-align: center;\">[latex](4x^3+12x^2) + (x+3)[\/latex]<\/p>\n<p>The GCF of the first group is [latex]4x^2[\/latex] so let&#8217;s factor [latex]4x^2[\/latex] out of the first group. What is the GCF of the second group? The common factor of the second pair is 1. There is no other factors that the terms have in common. We can write that as<\/p>\n<p style=\"text-align: center;\">[latex]4x^2(x+3)+ 1(x+3)[\/latex]<\/p>\n<p>Writing the [latex]1[\/latex] in front, makes the factoring more apparent. Factor out the common factor, [latex](x+3)[\/latex], from both terms, leaving [latex](4x^2+1)[\/latex] as the other factor.<\/p>\n<p style=\"text-align: center;\">[latex](x+3)(4x^2 +1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 id=\"Trinomial1\">Factoring Trinomials with Leading Coefficient of 1<\/h2>\n<p>Trinomials are polynomials with three terms. We are going to show you a method for factoring a trinomial whose leading coefficient is\u00a0[latex]1[\/latex]. \u00a0Although we should always begin by looking for a GCF, factoring out the GCF is not the only way that trinomials\u00a0can be factored. The trinomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of\u00a0[latex]1[\/latex], but it can be written as the product of the binomial factors.<\/p>\n<p>The last example in the previous section asked us to graph [latex]g(x)=x^2+5x+6[\/latex]. It appeared that it would be easier to find the [latex]x[\/latex]-intercept(s), vertex, and axis of symmetry if we could change the function to intercept form. How do we change this function from general form [latex]f(x)=ax^2+bx+c[\/latex] to intercept (factored) form [latex]f(x)=(x-p)(x-q)[\/latex]? We need to learn how to factor trinomials.<\/p>\n<p style=\"text-align: center;\">We will start by factoring the polynomial [latex]1x^2+5x+6[\/latex].<\/p>\n<p>Let&#8217;s identify the values for [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex]. Looking at the given polynomial, [latex]a=1[\/latex], [latex]b=5[\/latex], and [latex]c=6[\/latex].<\/p>\n<p>We need to find two integers [latex]m[\/latex] and [latex]n[\/latex] whose product is [latex]c[\/latex] and whose sum is [latex]b[\/latex]. In other words, let&#8217;s think of two integers that multiply to be [latex]6[\/latex] and add to be [latex]5[\/latex]. The factors of 6 are<\/p>\n<table style=\"width: 30%; font-size: 110%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, 6; -1, -6; -2, -3; and 2, 3. The entries in the second column are: 7, -7, -5, and 5.\">\n<thead>\n<tr>\n<th>Factors of [latex]6[\/latex]<\/th>\n<th>Sum of 5<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,6[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,-6[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,-3[\/latex]<\/td>\n<td>[latex]-5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #ffff00;\">[latex]2,3[\/latex]<\/span><\/td>\n<td><span style=\"background-color: #ffff00;\">[latex]5[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Which pair of factors add to be [latex]5[\/latex]? The only way to get a sum of 5 is by adding the [latex]2[\/latex] and [latex]3[\/latex] ([latex]2+3=5[\/latex]). We have identified [latex]m[\/latex] as [latex]2[\/latex] and [latex]n[\/latex] as [latex]3[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} ax^2+bx+c &= (x+m)(x+n)\\\\ &= (x+2)(x+3) \\end{align}[\/latex]<\/p>\n<p>NOTE: The answer could also be written as [latex](x+3)(x+2)[\/latex]. We can change the order of the factors since multiplication is commutative.<\/p>\n<p>To check if it is factored correctly, multiply the two binomials:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\left(x+2\\right)\\left(x+3\\right) &= x^2+3x+2x+6\\\\ &= x^2+5x+6 \\end{align}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>Factoring a Trinomial with Leading Coefficient 1<\/h3>\n<p>In general, for a trinomial of the form [latex]{x}^{2}+bx+c[\/latex], you can factor a trinomial with leading coefficient\u00a0[latex]1[\/latex] by finding two integers, [latex]m[\/latex] and [latex]n[\/latex], whose product is [latex]c[\/latex] and whose sum is [latex]b[\/latex].<\/p>\n<\/div>\n<p>Let us put this idea to practice with the following example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]{x}^{2}+2x - 15[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44696\">Show Solution<\/span><\/p>\n<div id=\"q44696\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with leading coefficient [latex]1, b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two integers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table style=\"width: 20%; font-size: 110%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\n<thead>\n<tr>\n<th>Factors of [latex]-15[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-15[\/latex]<\/td>\n<td>[latex]-14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,15[\/latex]<\/td>\n<td>[latex]14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-5[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #ffff00;\">[latex]-3,5[\/latex]<\/span><\/td>\n<td><span style=\"background-color: #ffff00;\">[latex]2[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now that we have identified [latex]m[\/latex] and [latex]n[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we present two more examples of factoring a trinomial with a leading coefficient of [latex]1[\/latex].<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/-SVBVVYVNTM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>To summarize our process, consider the following steps:<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\n<ol>\n<li>List factors of [latex]c[\/latex].<\/li>\n<li>Find [latex]m[\/latex] and [latex]n[\/latex], a pair of factors with a product of [latex]c[\/latex] and with a sum of [latex]b[\/latex].<\/li>\n<li>Write the factored expression as [latex]\\left(x+m\\right)\\left(x+n\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>In our next example, we show that when [latex]c[\/latex] is negative, either [latex]m[\/latex] or [latex]n[\/latex] will be negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]x^2+x-12[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q205737\">Show Solution<\/span><\/p>\n<div id=\"q205737\" class=\"hidden-answer\" style=\"display: none\">\n<p>Consider all the combinations of integers whose product is\u00a0[latex]-12[\/latex] and list their sum.<\/p>\n<table style=\"width: 30%; height: 84px; font-size: 110%;\">\n<thead>\n<tr style=\"height: 12px;\">\n<th style=\"height: 12px;\">Factors of [latex]\u221212[\/latex]<\/th>\n<th style=\"height: 12px;\">Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]1, \u221212[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]\u221211[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]2, \u22126[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]\u22124[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]3, \u22124[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]\u22121[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\"><span style=\"background-color: #ffff00;\">[latex]4, \u22123[\/latex]<\/span><\/td>\n<td style=\"height: 12px;\"><span style=\"background-color: #ffff00;\">[latex]1[\/latex]<\/span><\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]6, \u22122[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]4[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]12, \u22121[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]11[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Choose the values whose sum is\u00a0[latex]+1[\/latex]:\u00a0\u00a0[latex]m=4[\/latex] and [latex]n=\u22123[\/latex], and place them into a product of binomials.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]{x}^{2}-7x+6[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q662468\">Show Solution<\/span><\/p>\n<div id=\"q662468\" class=\"hidden-answer\" style=\"display: none\">\n<p>List the factors of\u00a0[latex]6[\/latex]. Note that the b term is negative, so we will need to consider negative numbers in our list.<\/p>\n<table style=\"width: 20%; font-size: 110%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\n<thead>\n<tr>\n<th>Factors of [latex]6[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,6[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2, 3[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #ffff00;\">[latex]-1, -6[\/latex]<\/span><\/td>\n<td><span style=\"background-color: #ffff00;\">[latex]-7[\/latex]<\/span><\/td>\n<\/tr>\n<tr>\n<td>[latex]-2, -3[\/latex]<\/td>\n<td>[latex]-5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Choose the pair that sum to [latex]-7[\/latex]: \u00a0[latex]m=-1[\/latex], [latex]n=-6[\/latex]<\/p>\n<p>Write the pair as constant terms in a product of binomials.<\/p>\n<p>[latex]\\left(x-1\\right)\\left(x-6\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the last example, the [latex]b[\/latex] term was negative and the [latex]c[\/latex] term was positive. This will always mean that if it can be factored, [latex]m[\/latex] and [latex]n[\/latex] will both be negative.<\/p>\n<h2 id=\"Trinomialnot1\">Factoring trinomials with Leading Coefficient not 1<\/h2>\n<p>Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by separating the [latex]x[\/latex] term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be factored as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as a 4-term polynomial [latex]2{x}^{2}+2x+3x+3[\/latex]. The process to do this is explained in the next example. We will \u00a0use factor by grouping to complete the factorization of the 4-term polynomial resulting in [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.<\/p>\n<div class=\"textbox\">\n<h3>AC-method of Factoring<\/h3>\n<p>To factor a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] by using the AC-Method, we find two integers with a product of [latex]a\\cdot c[\/latex] and a sum of [latex]b[\/latex]. We use these integers to separate the [latex]x[\/latex] term into the sum of two terms (making it a 4-term polynomial). Then factor the polynomial by grouping.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q806328\">Show Solution<\/span><\/p>\n<div id=\"q806328\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]a\\cdot c=-30[\/latex]. We need to find two integers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table style=\"width: 20%; height: 84px; font-size: 110%;\" summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\n<thead>\n<tr style=\"height: 12px;\">\n<th style=\"height: 12px;\">Factors of [latex]-30[\/latex]<\/th>\n<th style=\"height: 12px;\">Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]1,-30[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]-29[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]-1,30[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]29[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]2,-15[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]-13[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]-2,15[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]13[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\">[latex]3,-10[\/latex]<\/td>\n<td style=\"height: 12px;\">[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"height: 12px;\"><span style=\"background-color: #ffff00;\">[latex]-3,10[\/latex]<\/span><\/td>\n<td style=\"height: 12px;\"><span style=\"background-color: #ffff00;\">[latex]7[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So [latex]m=-3[\/latex] and [latex]n=10[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align}5{x}^{2}-3x+10x-6 && \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right) && \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\  \\left(5x - 3\\right)\\left(x+2\\right) && \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\n<div><\/div>\n<div>\n<p>We can check that the factorization is correct by multiplying the two binomials together.<\/p>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: AC-Method &#8211; Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex],<\/h3>\n<ol>\n<li>Factor out the GCF from all terms, if there is a GCF.<\/li>\n<li>List factors of [latex]a\\cdot c[\/latex].<\/li>\n<li>Find [latex]m[\/latex] and [latex]n[\/latex], a pair of factors of [latex]a\\cdot c[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Rewrite the original expression as a 4-term polynomial.<\/li>\n<li>Factor out the GCF of first two terms.<\/li>\n<li>Factor out the GCF of last two terms.<\/li>\n<li>Factor out the GCF of the expression.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]6x^2+11x+4[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q864920\">Show Solution<\/span><\/p>\n<div id=\"q864920\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with [latex]a=6,b=11[\/latex], and [latex]c=4[\/latex]. First, determine [latex]a\\cdot c=24[\/latex]. We need to find two numbers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]24[\/latex] and a sum of [latex]11[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table style=\"width: 20%; font-size: 110%;\" summary=\"A table with 9 rows and 2 columns. The first column is labeled: Factors of 24 while the second is labeled: Sum of Factors. The entries in the first column are: 1, 24; -1, -24; 2, 12; -2, -12; 3, 8; -3, -8; 4, 6; -4, -6. The entries in the second column are: 25, -25, 14, -14, 11, -11, 10 and -10.\">\n<thead>\n<tr>\n<th>Factors of [latex]24[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,24[\/latex]<\/td>\n<td>[latex]25[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,-24[\/latex]<\/td>\n<td>[latex]-25[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2,12[\/latex]<\/td>\n<td>[latex]14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,-12[\/latex]<\/td>\n<td>[latex]-14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><span style=\"background-color: #ffff00;\">[latex]3,8[\/latex]<\/span><\/td>\n<td><span style=\"background-color: #ffff00;\">[latex]11[\/latex]<\/span><\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,-8[\/latex]<\/td>\n<td>[latex]-11[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4,6[\/latex]<\/td>\n<td>[latex]10[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-4,-6[\/latex]<\/td>\n<td>[latex]-10[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So [latex]m=3[\/latex] and [latex]n=8[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align}6{x}^{2}+3x+8x+4 && \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\  3x\\left(2x +1\\right)+4\\left(2x +1\\right) && \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\ \\left(2x +1\\right)\\left(3x+4\\right) && \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\n<div><\/div>\n<div>\n<p>We can check that the factorization is correct by multiplying the two binomials together.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Factor completely. [latex]10x^2-7x-6[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q559688\">Show Solution<\/span><\/p>\n<div id=\"q559688\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with [latex]a=10,b=-7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]a\\cdot c=-60[\/latex]. We need to find two integers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]-60[\/latex] and a sum of [latex]-7[\/latex].<\/p>\n<p>You can begin by listing out all pairs that multiply to [latex]-60[\/latex] if you find this helpful.<\/p>\n<p>So [latex]m=-12[\/latex] and [latex]n=5[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align}10{x}^{2}-12x+5x-6 && \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\  2x\\left(5x-6\\right)+1\\left(5x-6\\right) && \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\ \\left(5x-6\\right)\\left(2x+1\\right) && \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\n<div><\/div>\n<div>\n<p>We can check that the factorization is correct by multiplying the two binomials together.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]5x^2-x+2[\/latex]<\/p>\n<p>We have a trinomial with [latex]a=5,b=-1[\/latex], and [latex]c=-2[\/latex]. First, determine [latex]a\\cdot c=10[\/latex]. We need to find two integers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]10[\/latex] and a sum of [latex]-1[\/latex].<\/p>\n<table style=\"width: 20%; font-size: 110%;\" summary=\"A table with 9 rows and 2 columns. The first column is labeled: Factors of 24 while the second is labeled: Sum of Factors. The entries in the first column are: 1, 24; -1, -24; 2, 12; -2, -12; 3, 8; -3, -8; 4, 6; -4, -6. The entries in the second column are: 25, -25, 14, -14, 11, -11, 10 and -10.\">\n<thead>\n<tr>\n<th>Factors of [latex]10[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,10[\/latex]<\/td>\n<td>[latex]11[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,-10[\/latex]<\/td>\n<td>[latex]-11[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2,5[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,-5[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We have exhausted all the possible integers that multiply to be [latex]10[\/latex] and none of the combinations of factors will add to be [latex]-1[\/latex]. Therefore, the polynomial is not factorable and is said to be <strong>prime<\/strong>.<\/p>\n<p>Answer: Not Factorable or Prime<\/p>\n<\/div>\n<p>The following video has another example about factoring trinomials when the leading coefficient is not 1.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/agDaQ_cZnNc?si=SqrqME5CdbK4p6O9\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In some situations, [latex]a[\/latex] is negative, as in [latex]-4h^2+11h+3[\/latex]. It often makes sense to factor out [latex]-1[\/latex] as the first step, making the polynomial easier to factor.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]-4h^2+11h+3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q745970\">Show Solution<\/span><\/p>\n<div id=\"q745970\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factor [latex]-1[\/latex] out of the trinomial. Notice that the signs of all three terms have changed.<\/p>\n<p style=\"text-align: center;\">[latex]-1(4h^2-11h-3)[\/latex]<\/p>\n<p>We now have a trinomial with [latex]a=4,b=-11[\/latex], and [latex]c=-3[\/latex]. First, determine [latex]a\\cdot c=-12[\/latex]. We need to find two numbers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]-12[\/latex] and a sum of [latex]-11[\/latex].<\/p>\n<p>You can begin by listing out all pairs that multiply to [latex]-12[\/latex] if you find this helpful.<\/p>\n<p>So [latex]m=-12[\/latex] and [latex]n=1[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align}-1(4{h}^{2}-12h+1h-3) && \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\  -1[\\,4h\\left(h-3\\right)+1\\left(h-3\\right)]\\, && \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\ -1\\left(h-3\\right)\\left(4h+1\\right) && \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\n<div><\/div>\n<div>\n<p>We can check that the factorization is correct by multiplying the two binomials together.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]16x^4-12x^3-10x^2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q945393\">Show Solution<\/span><\/p>\n<div id=\"q945393\" class=\"hidden-answer\" style=\"display: none\">\n<p>Is there a GCF that all three terms have in common? Yes, [latex]2x^2[\/latex] is a common factor of all three. We will factor this out first.<\/p>\n<p style=\"text-align: center;\">[latex]2x^2(8x^2-6x-5)[\/latex]<\/p>\n<p>The trinomial that we need to factor has coefficients of [latex]a=8,b=-6[\/latex], and [latex]c=-5[\/latex]. First, determine [latex]a\\cdot c=-40[\/latex]. We need to find two integers, [latex]m[\/latex] and [latex]n[\/latex], with a product of [latex]-40[\/latex] and a sum of [latex]-6[\/latex].<\/p>\n<p>You can begin by listing out all pairs that multiply to [latex]-40[\/latex] if you find this helpful.<\/p>\n<p>So [latex]m=-10[\/latex] and [latex]n=4[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\require{color}\\begin{align} 2x^2(8x^2-10x+4x-5) && \\color{blue}{\\textsf{Rewrite the original expression as a 4-term polynomial}}\\\\  2x^2[\\,2x\\left(4x-5\\right)+1\\left(4x-5\\right)]\\, && \\color{blue}{\\textsf{Factor out the GCF of each part}}\\\\ 2x^2\\left(4x-5\\right)\\left(2x+1\\right) && \\color{blue}{\\textsf{Factor out the GCF of the expression}}\\end{align}[\/latex]<\/div>\n<div><\/div>\n<div>\n<p>We can check that the factorization is correct by multiplying the two binomials together.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2 id=\"special\">Factoring Special Cases<\/h2>\n<h3>Factoring a Perfect Square Trinomial<\/h3>\n<p>What is a perfect square? A perfect square is the product of an integer multiplied by itself. For example, [latex]4 \\cdot 4 = 16[\/latex] So, [latex]16[\/latex] is a perfect square. Other examples of perfect squares are, [latex]1, 4, 9, 16, 25, 36, 49, 64, 81[\/latex], and [latex]100[\/latex]. A perfect square trinomial is a trinomial that can be written in factored form as a binomial squared. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}{\\left(a+b\\right)}^{2}\\hfill & =& {a}^{2}+2ab+{b}^{2}\\hfill \\\\ & \\text{and}& \\\\ \\hfill {\\left(a-b\\right)}^{2}& =& {a}^{2}-2ab+{b}^{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: left;\">We can use these same equations to factor any perfect square trinomial.<\/div>\n<div class=\"textbox\">\n<h3>Perfect Square Trinomials<\/h3>\n<p>A perfect square trinomial can be written, in factored form, as a binomial squared:<\/p>\n<div style=\"text-align: center;\">[latex]\\large{{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]\\large{{a}^{2}-2ab+{b}^{2}={\\left(a-b\\right)}^{2}}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]25{x}^{2}+20x+4[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q114092\">Show Solution<\/span><\/p>\n<div id=\"q114092\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, notice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex].<\/p>\n<p>In using the equations above to help you factor this, we can say that [latex]a=5x[\/latex] and [latex]b=2[\/latex].<\/p>\n<p>Next, check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex].<\/p>\n<p>Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{(a+b)}^2={\\left(5x+2\\right)}^{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a perfect square trinomial, factor it into the square of a binomial<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Confirm that the middle term is twice the product of [latex]a\\cdot b[\/latex].<\/li>\n<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex] or [latex]{(a-b)}^{2}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>In the following video, we provide another short description of perfect square trinomials and shows how to factor them using a formula.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/UMCVGDTxxTI?si=yqGYn8z1CD59WvPn\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Factoring a Difference of Squares<\/h3>\n<p>A difference of squares can be factored into two binomials but has only two terms. \u00a0Let\u2019s start from the product of two special binomials to see the pattern.<\/p>\n<p>Consider the product of the following two binomials, which are identical except for the operation: [latex](x-2)(x+2)[\/latex]. If we multiply them together, the middle term zeros out.<\/p>\n<p style=\"text-align: center;\">Multiply. [latex](x-2)(x+2)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &= x^2-2x+2x-2^2\\\\ &= x^2-2^2\\\\ &= x^2-4 \\end{align}[\/latex]<\/p>\n<p>The polynomial [latex]x^2-4[\/latex] is called a difference of squares because each term is a perfect square and the operation connecting them is subtraction. A difference of squares will always factor in the following way:<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<p>[latex]\\\\[\/latex]<\/p>\n<p>We can use this equation to factor any differences of squares.<\/p>\n<div class=\"textbox\">\n<h3>Differences of Squares<\/h3>\n<p>A difference of squares can be rewritten as two factors containing the same terms but opposite signs.<\/p>\n<div style=\"text-align: center;\">[latex]\\large{{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)}[\/latex]<\/div>\n<\/div>\n<div>The following video shows examples of factoring a difference of squares.<\/div>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/Li9IBp5HrFA?si=sPSn9f6DDEFLsmlw\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div><\/div>\n<div class=\"textbox\">\n<h3>How To: Given a difference of squares, factor it into binomials<\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Determine [latex]a[\/latex] and [latex]b[\/latex].<\/li>\n<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely. [latex]9{x}^{2}-25[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q830417\">Show Solution<\/span><\/p>\n<div id=\"q830417\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex].<\/p>\n<p>This means that [latex]a=3x[\/latex] and [latex]b=5[\/latex].<\/p>\n<p>The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].<\/p>\n<p>You can <strong>check<\/strong> that it is factored correctly by multiplying the two binomials.<\/p>\n<p>[latex]\\begin{align}(3x+5)(3x-5)\\\\ &= 9x^2-15x+15x-5^2\\\\ &= 9x^2-5^2\\\\ &= 9x^2-25 \\end{align}[\/latex]<\/p>\n<p>Multiplying the binomials results in the polynomial that we started with so, it is indeed factored correctly. Our factored answer is:<\/p>\n<p style=\"text-align: center;\">[latex](3x+5)(3x-5)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Factor completely. [latex]18x^2-2x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q633557\">Show Solution<\/span><\/p>\n<div id=\"q633557\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that the terms of this binomial are not perfect squares. However, they do share a common factor of [latex]2x[\/latex] which can be factored out.<\/p>\n<p style=\"text-align: center;\">[latex]2x(9x^2-1)[\/latex]<\/p>\n<p>The resulting binomial is now a difference of squares. Since [latex]9x^2=(3x)^2[\/latex] and [latex]1=1^2[\/latex], we have [latex]a=3x[\/latex] and [latex]b=1[\/latex]. Applying our difference of squares equation, and remembering the fact of [latex]2x[\/latex] in front, this factors as<\/p>\n<p style=\"text-align: center;\">[latex]2x(3x+1)(3x-1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Summary<\/h2>\n<p>In all factoring problems, we first look for a common factor that can be factored out of all terms. After achieving this (if possible), we have now learned special factoring techniques in this section.<\/p>\n<p>If a trinomial is of the form of a perfect square trinomial, it factors as<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\n<div style=\"text-align: center;\">or<\/div>\n<div style=\"text-align: center;\">[latex]{a}^{2}-2ab+{b}^{2}={\\left(a-b\\right)}^{2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div>If a binomial is of the form of a difference of squares, [latex]a^2-b^2[\/latex], it will factor as<\/div>\n<div style=\"text-align: center;\">[latex]a^2-b^2= (a+b)(a-b)[\/latex]<\/div>\n<div><\/div>\n<div>If a binomial is of the form of a sum of squares, [latex]a^2+b^2[\/latex], it will not factor and is considered a prime polynomial.<\/div>\n<div><\/div>\n<div><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-159\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 1: Identify GCF and Factor a Binomial. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/25_f_mVab_4\">https:\/\/youtu.be\/25_f_mVab_4<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Identify GCF and Factor a Trinomial. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/3f1RFTIw2Ng\">https:\/\/youtu.be\/3f1RFTIw2Ng<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex 1: Identify GCF and Factor a Binomial\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/25_f_mVab_4\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 12: Factoring, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Identify GCF and Factor a Trinomial\",\"author\":\"James Sousa (Mathispower4u.com) 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