{"id":2714,"date":"2016-07-18T16:24:13","date_gmt":"2016-07-18T16:24:13","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2714"},"modified":"2018-05-16T23:45:00","modified_gmt":"2018-05-16T23:45:00","slug":"read-the-greatest-common-factor","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/read-the-greatest-common-factor\/","title":{"raw":"Factoring Trinomials","rendered":"Factoring Trinomials"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Greatest common factor\r\n<ul>\r\n \t<li>Identify the difference between a factor and the act of factoring<\/li>\r\n \t<li>Identify the greatest common factor of a polynomial<\/li>\r\n \t<li>Factor the greatest common factor out of\u00a0a polynomial<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Factor a Trinomial with Leading Coefficient = 1\r\n<ul>\r\n \t<li>Identify a trinomial<\/li>\r\n \t<li>Identify the leading coefficient of a trinomial<\/li>\r\n \t<li>Use a method to factor a trinomial with a leading coefficient of 1<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Factor by Grouping\r\n<ul>\r\n \t<li>Factor a trinomial with\u00a0leading coefficient other than 1\u00a0using grouping<\/li>\r\n \t<li>Recognize when a trinomial cannot be factored<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<strong>Factors<\/strong> are the building blocks of multiplication. They are the numbers that you can multiply together to produce another number: 2 and 10 are factors of 20, as are 4, 5, 1, 20. To factor a number is to rewrite it as a product. [latex]<span class=\"katex\"><span class=\"katex-mathml\">\\displaystyle 20=4\\cdot5<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathrm\">2<\/span><span class=\"mord mathrm\">0<\/span><span class=\"mrel\">=<\/span><span class=\"mord mathrm\">4<\/span><span class=\"mbin\">\u22c5<\/span><span class=\"mord mathrm\">5[\/latex]<\/span><\/span><\/span><\/span><\/span>. 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 numbers is the largest number that divides evenly into both numbers. For instance, [latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest number 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\nWhen factoring a polynomial expression, our first step should be to check for a GCF. 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 the largest polynomial that divides evenly into the polynomials.\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}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\text{GCF}=5b^{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe monomials have the factors 5, <i>b<\/i>, and <i>b<\/i> 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}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\text{GCF}=9c^{2}d[\/latex]\r\n\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\nNow that you have practiced identifying the GCF of a term 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 a factored form. Recall that the <strong>distributive property of multiplication over addition<\/strong> states that a product of a number and a sum is the same as the sum of the products.\r\n<div class=\"textbox shaded\">\r\n<h4>Distributive Property Forward and Backward<\/h4>\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<em>a<\/em> 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\"]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].\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\r\n<h4>Answer<\/h4>\r\n[latex]5b^{2}\\left(5b+2\\right)[\/latex]\r\n\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\r\nWe will show one last example of finding the GCF of a polynomial with several terms and two variables. No matter how large the polynomial, you can use the same technique described below to factor out it's GCF.\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>Combine 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\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].\r\n[reveal-answer q=\"112050\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"112050\"]\r\n\r\nFirst, find the GCF of the expression. 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 exponent of lowest degree.) And the GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. Combine these 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<div>[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Analysis of the Solution<\/h3>\r\nAfter factoring, we can check our work by multiplying. Use the distributive property to confirm that [latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].\r\n\r\nIn the following video you will see two more\u00a0example of how to find and factor our the greatest common factor of a polynomial.\r\n\r\nhttps:\/\/youtu.be\/3f1RFTIw2Ng\r\n<h2>Factor a Trinomial with Leading Coefficient = 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 1. \u00a0Although we should always begin by looking for a GCF, pulling 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 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].\r\n\r\nRecall how to use the distributive property to multiply two binomials:\r\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(x+3\\right) = x^2+3x+2x+6=x^2+5x+6[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We can reverse the distributive property and return [latex]x^2+5x+6\\text{ to }\\left(x+2\\right)\\left(x+3\\right) [\/latex]\u00a0by finding two numbers with a product of [latex]6[\/latex] and a sum of [latex]5[\/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 1 by finding two numbers,[latex]p[\/latex] and [latex]q[\/latex]\u00a0whose product is c, and whose sum is b.\r\n\r\n<\/div>\r\nLet's put this idea to practice with the following example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]{x}^{2}+2x - 15[\/latex].\r\n[reveal-answer q=\"44696\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"44696\"]We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers 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%;\" 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>14<\/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>[latex]-3,5[\/latex]<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow that we have identified [latex]p[\/latex] and [latex]q[\/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 1.\r\n\r\nhttps:\/\/youtu.be\/-SVBVVYVNTM\r\n\r\nTo summarize our process consider these 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]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\r\n<\/ol>\r\nWe will now show an example where the trinomial has a negative c term. Pay attention to the signs of the numbers that are considered for p and q.\r\n\r\n<\/div>\r\nIn our next example, we show that when c is negative, either p or q will be negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[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 numbers whose product is -12, and list their sum.\r\n<table style=\"width: 30%;\">\r\n<thead>\r\n<tr>\r\n<th>Factors whose product is [latex]\u221212[\/latex]<\/th>\r\n<th>Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1\\cdot\u221212=\u221212[\/latex]<\/td>\r\n<td>[latex]1+\u221212=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2\\cdot\u22126=\u221212[\/latex]<\/td>\r\n<td>[latex]2+\u22126=\u22124[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\\cdot\u22124=\u221212[\/latex]<\/td>\r\n<td>[latex]3+\u22124=\u22121[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4\\cdot\u22123=\u221212[\/latex]<\/td>\r\n<td>[latex]4+\u22123=1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6\\cdot\u22122=\u221212[\/latex]<\/td>\r\n<td>[latex]6+\u22122=4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]12\\cdot\u22121=\u221212[\/latex]<\/td>\r\n<td>[latex]12+\u22121=11[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nChoose the values whose sum is +1:\u00a0\u00a0[latex]r=4[\/latex] and [latex]s=\u22123[\/latex], and place them into a product of binomials.\r\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nWhich property of multiplication can be used to describe why\u00a0[latex]\\left(x+4\\right)\\left(x-3\\right) =\\left(x-3\\right)\\left(x+4\\right)[\/latex]. Use the textbox below to write down your ideas before you look at the answer.\r\n\r\n[practice-area rows=\"2\"][\/practice-area]\r\n[reveal-answer q=\"177955\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"177955\"]\r\n\r\nThe <strong>commutative property of multiplication<\/strong> states that numbers may be multiplied in any order without affecting the product.\r\n<div class=\"bcc-box bcc-success\">\r\n<div style=\"text-align: center;\">[latex]a\\cdot b=b\\cdot a[\/latex]<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last example we will show how to factor a trinomial whose b term is negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]{x}^{2}-7x+6[\/latex].\r\n[reveal-answer q=\"662468\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"662468\"]\r\n\r\nList the factors of 6. Note that the b term is negative - so we will need to consider negative numbers in our list.\r\n<table style=\"width: 20%;\" 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>[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<\/tbody>\r\n<\/table>\r\nChoose the pair that sum to [latex]-7[\/latex], which is\u00a0[latex]-1, -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-7\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>\u00a0Analysis of the solution<\/h3>\r\nIn the last example, the b\u00a0term was negative and the c term was positive. This will always mean that if it can be factored, p and q\u00a0will both be negative.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nCan every trinomial be factored as a product of binomials?\r\n\r\nMathematicians often use a counter example to prove\u00a0or disprove a question. A counter example means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient 1 that\u00a0<em>cannot\u00a0<\/em>be factored as a product of binomials?\r\n\r\nUse the textbox below to write your ideas.\r\n\r\n[practice-area rows=\"2\"][\/practice-area]\r\n[reveal-answer q=\"776075\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"776075\"]Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime.\r\n\r\nA counter-example would be: [latex]x^2+3x+7[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Factor by Grouping<\/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 dividing the <em>x<\/em> 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 rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [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\r\nThe first step in this process is to figure out what two numbers to use to re-write the\u00a0<em>x<\/em> term as the sum of two new terms. Making a table to keep track of your work is helpful. We are looking for two numbers with a product of [latex]2\\cdot3=6[\/latex] and a sum of [latex]5[\/latex]\r\n<table class=\" aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 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]2\\cdot3=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]-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>[latex]-2,-3[\/latex]<\/td>\r\n<td>[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe pair [latex]p=2,\\text{ and }q=3[\/latex] will give the correct\u00a0<em>x<\/em> term, so we will rewrite it using the new factors:\r\n<p style=\"text-align: center;\">[latex]2{x}^{2}+5x+3=2x^2+2x+3x+3[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we can group the polynomial into two binomials.<\/p>\r\n<p style=\"text-align: center;\">[latex]2x^2+2x+3x+3=(2x^2+2)+(3x+3)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Identify the GCF of each binomial.<\/p>\r\n<p style=\"text-align: left;\">2x is the GCF of [latex](2x^2+2)[\/latex] and 3 is the GCF of [latex](3x+3)[\/latex], use this to rewrite the polynomial:<\/p>\r\n<p style=\"text-align: center;\">[latex](2x^2+2x)+(3x+3)=2x(x+1)+3(x+1)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Note how we leave the signs in the binomials and the addition that joins them, be careful with signs when you factor out the GCF. The GCF of our new polynomial is [latex](x+1)[\/latex], we factor this out as well:<\/p>\r\n<p style=\"text-align: center;\">[latex]2x(x+1)+3(x+1)=(x+1)(2x+3)[\/latex].<\/p>\r\n<p style=\"text-align: left;\">Sometimes it helps visually to write the polynomial this way\u00a0[latex](x+1)2x+(x+1)3[\/latex] before you factor out the GCF. This is purely a matter of preference, multiplication is commutative, so order doesn't matter.<\/p>\r\n\r\n<div class=\"textbox\" style=\"text-align: center;\">\r\n<h3 style=\"text-align: left;\">\u00a0A General Note: Factor by Grouping<\/h3>\r\n<p style=\"text-align: left;\">To factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.<\/p>\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[reveal-answer q=\"658545\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"658545\"]\r\n\r\nWe have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers 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 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>\r\n<th>Factors of [latex]-30[\/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,-30[\/latex]<\/td>\r\n<td>[latex]-29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,30[\/latex]<\/td>\r\n<td>29<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2,-15[\/latex]<\/td>\r\n<td>[latex]-13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,15[\/latex]<\/td>\r\n<td>13<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-10[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,10[\/latex]<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]p=-3[\/latex] and [latex]q=10[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill &amp; \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill &amp; \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill &amp; \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<h3>Analysis of the Solution<\/h3>\r\nWe can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].\r\n\r\n<\/div>\r\nWe can summarize our process in the following way:\r\n<h3>Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping.<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]ac[\/latex].<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]ac[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\r\n \t<li>Pull out the GCF of [latex]a{x}^{2}+px[\/latex].<\/li>\r\n \t<li>Pull out the GCF of [latex]qx+c[\/latex].<\/li>\r\n \t<li>Factor out the GCF of the expression.<\/li>\r\n<\/ol>\r\nIn the following video we present one more example of factoring a trinomial whose leading coefficient is not 1 using the grouping method.\r\n\r\nhttps:\/\/youtu.be\/agDaQ_cZnNc\r\n\r\nFactoring trinomials whose leading coefficient is not 1 becomes quick and kind of fun once you get the idea. \u00a0Give the next example a try on your own before you look at the solution.\r\n<div style=\"text-align: center;\">\r\n<p style=\"text-align: left;\">We will show two more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]2{x}^{2}+9x+9[\/latex].\r\n[reveal-answer q=\"834453\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"834453\"]\r\n\r\nFind two numbers p, q such that [latex]p\\cdot{q}=18[\/latex], and [latex]p + q = 9[\/latex]. 9 and 18 are both positive, so we will only consider positive factors.\r\n<table class=\" aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 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]2\\cdot9=18[\/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, 18[\/latex]<\/td>\r\n<td>[latex]19[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,6[\/latex]<\/td>\r\n<td>[latex]9[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can stop because we have found our factors.\r\n\r\nRewrite the original expression, and group.\r\n<p style=\"text-align: center;\">[latex]2x^2+3x+6x+9=(2x^2+3x)+(6x+9)[\/latex]<\/p>\r\nFactor out the GCF of each binomial, and write as a product of two binomials:\r\n<p style=\"text-align: center;\">[latex](2x^2+3x)+(6x+9)=x(2x+3)+3(2x+3)=(x+3)(2x+3)[\/latex]<\/p>\r\n[latex]2{x}^{2}+9x+9=(x+3)(2x+3)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is an\u00a0example where the x term is positive and c is negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]6{x}^{2}+x - 1[\/latex].\r\n[reveal-answer q=\"806715\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"806715\"]\r\n<table class=\" aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 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\\cdot-1=-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]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1,-6[\/latex]<\/td>\r\n<td>[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,3[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can stop because we have found our factors.\r\n\r\nRewrite the original expression, and group.\r\n<p style=\"text-align: center;\">[latex]6{x}^{2}+x - 1=6x^2-2x+3x-1[\/latex]<\/p>\r\nFactor out the GCF of each binomial, and write as a product of two binomials:\r\n<p style=\"text-align: center;\">[latex](6x^2-2x)+(3x-1)=2x(3x-1)+1(3x-1)=(2x+1)(3x-1)[\/latex]<\/p>\r\n[latex]6{x}^{2}+x - 1=(2x+1)(3x-1)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video example, we will factor a trinomial whose leading term is negative.\r\n\r\nhttps:\/\/youtu.be\/zDAMjdBfkDs\r\n\r\nFor our last example, you will see that sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.\r\n<div class=\"bcc-box bcc-info\" style=\"text-align: left;\">\r\n<h3>Example<\/h3>\r\nFactor [latex]7x^{2}-16x-5[\/latex].\r\n[reveal-answer q=\"262926\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"262926\"]Find [latex]p, q[\/latex] such that [latex]p\\cdot{q}=-35\\text{ and }p+q=-16[\/latex]\r\n<table class=\" aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 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]2\\cdot3=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, 35[\/latex]<\/td>\r\n<td>[latex]34[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1, -35[\/latex]<\/td>\r\n<td>[latex]-34[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-5, 7[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-7,5[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Answer<\/h4>\r\nCannot be factored\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Greatest common factor\n<ul>\n<li>Identify the difference between a factor and the act of factoring<\/li>\n<li>Identify the greatest common factor of a polynomial<\/li>\n<li>Factor the greatest common factor out of\u00a0a polynomial<\/li>\n<\/ul>\n<\/li>\n<li>Factor a Trinomial with Leading Coefficient = 1\n<ul>\n<li>Identify a trinomial<\/li>\n<li>Identify the leading coefficient of a trinomial<\/li>\n<li>Use a method to factor a trinomial with a leading coefficient of 1<\/li>\n<\/ul>\n<\/li>\n<li>Factor by Grouping\n<ul>\n<li>Factor a trinomial with\u00a0leading coefficient other than 1\u00a0using grouping<\/li>\n<li>Recognize when a trinomial cannot be factored<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p><strong>Factors<\/strong> are the building blocks of multiplication. They are the numbers that you can multiply together to produce another number: 2 and 10 are factors of 20, as are 4, 5, 1, 20. To factor a number is to rewrite it as a product. [latex]<span class=\"katex\"><span class=\"katex-mathml\">\\displaystyle 20=4\\cdot5<\/span><span class=\"katex-html\"><span class=\"base textstyle uncramped\"><span class=\"reset-textstyle displaystyle textstyle uncramped\"><span class=\"mord mathrm\">2<\/span><span class=\"mord mathrm\">0<\/span><span class=\"mrel\">=<\/span><span class=\"mord mathrm\">4<\/span><span class=\"mbin\">\u22c5<\/span><span class=\"mord mathrm\">5[\/latex]<\/span><\/span><\/span><\/span><\/span>. 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 numbers is the largest number that divides evenly into both numbers. For instance, [latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest number 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>When factoring a polynomial expression, our first step should be to check for a GCF. 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 the largest polynomial that divides evenly into the polynomials.<\/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}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\text{GCF}=5b^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The monomials have the factors 5, <i>b<\/i>, and <i>b<\/i> 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}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\text{GCF}=9c^{2}d[\/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>Now that you have practiced identifying the GCF of a term 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 a factored form. Recall that the <strong>distributive property of multiplication over addition<\/strong> states that a product of a number and a sum is the same as the sum of the products.<\/p>\n<div class=\"textbox shaded\">\n<h4>Distributive Property Forward and Backward<\/h4>\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<em>a<\/em> 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\">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<h4>Answer<\/h4>\n<p>[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<p>We will show one last example of finding the GCF of a polynomial with several terms and two variables. No matter how large the polynomial, you can use the same technique described below to factor out it&#8217;s GCF.<\/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>Combine 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>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [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 Answer<\/span><\/p>\n<div id=\"q112050\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, find the GCF of the expression. 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 exponent of lowest degree.) And the GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. Combine these 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<div>[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3>Analysis of the Solution<\/h3>\n<p>After factoring, we can check our work by multiplying. Use the distributive property to confirm that [latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].<\/p>\n<p>In the following video you will see two more\u00a0example of how to find and factor our 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<h2>Factor a Trinomial with Leading Coefficient = 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 1. \u00a0Although we should always begin by looking for a GCF, pulling 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 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].<\/p>\n<p>Recall how to use the distributive property to multiply two binomials:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(x+3\\right) = x^2+3x+2x+6=x^2+5x+6[\/latex]<\/p>\n<p style=\"text-align: left;\">We can reverse the distributive property and return [latex]x^2+5x+6\\text{ to }\\left(x+2\\right)\\left(x+3\\right)[\/latex]\u00a0by finding two numbers with a product of [latex]6[\/latex] and a sum of [latex]5[\/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 1 by finding two numbers,[latex]p[\/latex] and [latex]q[\/latex]\u00a0whose product is c, and whose sum is b.<\/p>\n<\/div>\n<p>Let&#8217;s put this idea to practice with the following example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [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 Answer<\/span><\/p>\n<div id=\"q44696\" class=\"hidden-answer\" style=\"display: none\">We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers 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%;\" 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>14<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-5[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,5[\/latex]<\/td>\n<td>2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now that we have identified [latex]p[\/latex] and [latex]q[\/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 1.<\/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 these 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]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\n<\/ol>\n<p>We will now show an example where the trinomial has a negative c term. Pay attention to the signs of the numbers that are considered for p and q.<\/p>\n<\/div>\n<p>In our next example, we show that when c is negative, either p or q will be negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[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 numbers whose product is -12, and list their sum.<\/p>\n<table style=\"width: 30%;\">\n<thead>\n<tr>\n<th>Factors whose product is [latex]\u221212[\/latex]<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot\u221212=\u221212[\/latex]<\/td>\n<td>[latex]1+\u221212=\u221211[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2\\cdot\u22126=\u221212[\/latex]<\/td>\n<td>[latex]2+\u22126=\u22124[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\\cdot\u22124=\u221212[\/latex]<\/td>\n<td>[latex]3+\u22124=\u22121[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4\\cdot\u22123=\u221212[\/latex]<\/td>\n<td>[latex]4+\u22123=1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]6\\cdot\u22122=\u221212[\/latex]<\/td>\n<td>[latex]6+\u22122=4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]12\\cdot\u22121=\u221212[\/latex]<\/td>\n<td>[latex]12+\u22121=11[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Choose the values whose sum is +1:\u00a0\u00a0[latex]r=4[\/latex] and [latex]s=\u22123[\/latex], and place them into a product of binomials.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Which property of multiplication can be used to describe why\u00a0[latex]\\left(x+4\\right)\\left(x-3\\right) =\\left(x-3\\right)\\left(x+4\\right)[\/latex]. Use the textbox below to write down your ideas before you look at the answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"2\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q177955\">Show Answer<\/span><\/p>\n<div id=\"q177955\" class=\"hidden-answer\" style=\"display: none\">\n<p>The <strong>commutative property of multiplication<\/strong> states that numbers may be multiplied in any order without affecting the product.<\/p>\n<div class=\"bcc-box bcc-success\">\n<div style=\"text-align: center;\">[latex]a\\cdot b=b\\cdot a[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In our last example we will show how to factor a trinomial whose b term is negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [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 Answer<\/span><\/p>\n<div id=\"q662468\" class=\"hidden-answer\" style=\"display: none\">\n<p>List the factors of 6. Note that the b term is negative &#8211; so we will need to consider negative numbers in our list.<\/p>\n<table style=\"width: 20%;\" 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>[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<\/tbody>\n<\/table>\n<p>Choose the pair that sum to [latex]-7[\/latex], which is\u00a0[latex]-1, -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-7\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>\u00a0Analysis of the solution<\/h3>\n<p>In the last example, the b\u00a0term was negative and the c term was positive. This will always mean that if it can be factored, p and q\u00a0will both be negative.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Can every trinomial be factored as a product of binomials?<\/p>\n<p>Mathematicians often use a counter example to prove\u00a0or disprove a question. A counter example means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient 1 that\u00a0<em>cannot\u00a0<\/em>be factored as a product of binomials?<\/p>\n<p>Use the textbox below to write your ideas.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"2\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q776075\">Show Answer<\/span><\/p>\n<div id=\"q776075\" class=\"hidden-answer\" style=\"display: none\">Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/p>\n<p>A counter-example would be: [latex]x^2+3x+7[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Factor by Grouping<\/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 dividing the <em>x<\/em> 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 rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [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<p>The first step in this process is to figure out what two numbers to use to re-write the\u00a0<em>x<\/em> term as the sum of two new terms. Making a table to keep track of your work is helpful. We are looking for two numbers with a product of [latex]2\\cdot3=6[\/latex] and a sum of [latex]5[\/latex]<\/p>\n<table class=\"aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 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]2\\cdot3=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]-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>[latex]-2,-3[\/latex]<\/td>\n<td>[latex]-5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The pair [latex]p=2,\\text{ and }q=3[\/latex] will give the correct\u00a0<em>x<\/em> term, so we will rewrite it using the new factors:<\/p>\n<p style=\"text-align: center;\">[latex]2{x}^{2}+5x+3=2x^2+2x+3x+3[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we can group the polynomial into two binomials.<\/p>\n<p style=\"text-align: center;\">[latex]2x^2+2x+3x+3=(2x^2+2)+(3x+3)[\/latex]<\/p>\n<p style=\"text-align: left;\">Identify the GCF of each binomial.<\/p>\n<p style=\"text-align: left;\">2x is the GCF of [latex](2x^2+2)[\/latex] and 3 is the GCF of [latex](3x+3)[\/latex], use this to rewrite the polynomial:<\/p>\n<p style=\"text-align: center;\">[latex](2x^2+2x)+(3x+3)=2x(x+1)+3(x+1)[\/latex]<\/p>\n<p style=\"text-align: left;\">Note how we leave the signs in the binomials and the addition that joins them, be careful with signs when you factor out the GCF. The GCF of our new polynomial is [latex](x+1)[\/latex], we factor this out as well:<\/p>\n<p style=\"text-align: center;\">[latex]2x(x+1)+3(x+1)=(x+1)(2x+3)[\/latex].<\/p>\n<p style=\"text-align: left;\">Sometimes it helps visually to write the polynomial this way\u00a0[latex](x+1)2x+(x+1)3[\/latex] before you factor out the GCF. This is purely a matter of preference, multiplication is commutative, so order doesn&#8217;t matter.<\/p>\n<div class=\"textbox\" style=\"text-align: center;\">\n<h3 style=\"text-align: left;\">\u00a0A General Note: Factor by Grouping<\/h3>\n<p style=\"text-align: left;\">To factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.<\/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=\"q658545\">Show Answer<\/span><\/p>\n<div id=\"q658545\" 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]ac=-30[\/latex]. We need to find two numbers 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 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>\n<th>Factors of [latex]-30[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-30[\/latex]<\/td>\n<td>[latex]-29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,30[\/latex]<\/td>\n<td>29<\/td>\n<\/tr>\n<tr>\n<td>[latex]2,-15[\/latex]<\/td>\n<td>[latex]-13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,15[\/latex]<\/td>\n<td>13<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-10[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,10[\/latex]<\/td>\n<td>7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So [latex]p=-3[\/latex] and [latex]q=10[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill & \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill & \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill & \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<h3>Analysis of the Solution<\/h3>\n<p>We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].<\/p>\n<\/div>\n<p>We can summarize our process in the following way:<\/p>\n<h3>Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping.<\/h3>\n<ol>\n<li>List factors of [latex]ac[\/latex].<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]ac[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\n<li>Pull out the GCF of [latex]a{x}^{2}+px[\/latex].<\/li>\n<li>Pull out the GCF of [latex]qx+c[\/latex].<\/li>\n<li>Factor out the GCF of the expression.<\/li>\n<\/ol>\n<p>In the following video we present one more example of factoring a trinomial whose leading coefficient is not 1 using the grouping method.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/agDaQ_cZnNc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Factoring trinomials whose leading coefficient is not 1 becomes quick and kind of fun once you get the idea. \u00a0Give the next example a try on your own before you look at the solution.<\/p>\n<div style=\"text-align: center;\">\n<p style=\"text-align: left;\">We will show two more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]2{x}^{2}+9x+9[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834453\">Show Answer<\/span><\/p>\n<div id=\"q834453\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find two numbers p, q such that [latex]p\\cdot{q}=18[\/latex], and [latex]p + q = 9[\/latex]. 9 and 18 are both positive, so we will only consider positive factors.<\/p>\n<table class=\"aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 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]2\\cdot9=18[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1, 18[\/latex]<\/td>\n<td>[latex]19[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,6[\/latex]<\/td>\n<td>[latex]9[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We can stop because we have found our factors.<\/p>\n<p>Rewrite the original expression, and group.<\/p>\n<p style=\"text-align: center;\">[latex]2x^2+3x+6x+9=(2x^2+3x)+(6x+9)[\/latex]<\/p>\n<p>Factor out the GCF of each binomial, and write as a product of two binomials:<\/p>\n<p style=\"text-align: center;\">[latex](2x^2+3x)+(6x+9)=x(2x+3)+3(2x+3)=(x+3)(2x+3)[\/latex]<\/p>\n<p>[latex]2{x}^{2}+9x+9=(x+3)(2x+3)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here is an\u00a0example where the x term is positive and c is negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]6{x}^{2}+x - 1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q806715\">Show Answer<\/span><\/p>\n<div id=\"q806715\" class=\"hidden-answer\" style=\"display: none\">\n<table class=\"aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 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\\cdot-1=-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]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1,-6[\/latex]<\/td>\n<td>[latex]-5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,3[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We can stop because we have found our factors.<\/p>\n<p>Rewrite the original expression, and group.<\/p>\n<p style=\"text-align: center;\">[latex]6{x}^{2}+x - 1=6x^2-2x+3x-1[\/latex]<\/p>\n<p>Factor out the GCF of each binomial, and write as a product of two binomials:<\/p>\n<p style=\"text-align: center;\">[latex](6x^2-2x)+(3x-1)=2x(3x-1)+1(3x-1)=(2x+1)(3x-1)[\/latex]<\/p>\n<p>[latex]6{x}^{2}+x - 1=(2x+1)(3x-1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video example, we will factor a trinomial whose leading term is negative.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Factor a Trinomial in the Form -ax^2+bx+c Using the Grouping Technique\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/zDAMjdBfkDs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>For our last example, you will see that sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.<\/p>\n<div class=\"bcc-box bcc-info\" style=\"text-align: left;\">\n<h3>Example<\/h3>\n<p>Factor [latex]7x^{2}-16x-5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q262926\">Show Solution<\/span><\/p>\n<div id=\"q262926\" class=\"hidden-answer\" style=\"display: none\">Find [latex]p, q[\/latex] such that [latex]p\\cdot{q}=-35\\text{ and }p+q=-16[\/latex]<\/p>\n<table class=\"aligncenter\" style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 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]2\\cdot3=6[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]-1, 35[\/latex]<\/td>\n<td>[latex]34[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1, -35[\/latex]<\/td>\n<td>[latex]-34[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-5, 7[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-7,5[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Answer<\/h4>\n<p>Cannot be factored<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\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-2714\">\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><li>Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/-SVBVVYVNTM\">https:\/\/youtu.be\/-SVBVVYVNTM<\/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 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><li>Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/agDaQ_cZnNc\">https:\/\/youtu.be\/agDaQ_cZnNc<\/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>Factor a Trinomial in the Form -ax^2+bx+c Using the Grouping Technique. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/zDAMjdBfkDs\">https:\/\/youtu.be\/zDAMjdBfkDs<\/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":21,"menu_order":6,"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) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/3f1RFTIw2Ng\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Factor a Trinomial Using the Shortcut Method - 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