{"id":16323,"date":"2019-10-03T03:21:55","date_gmt":"2019-10-03T03:21:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-factor-by-grouping\/"},"modified":"2020-10-22T09:32:19","modified_gmt":"2020-10-22T09:32:19","slug":"read-factor-by-grouping","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/chapter\/read-factor-by-grouping\/","title":{"raw":"13.2.b - Factoring by Grouping","rendered":"13.2.b &#8211; Factoring by Grouping"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcome<\/h3>\r\n<ul>\r\n \t<li>Apply an algorithm to rewrite a trinomial as a four term polynomial and factor<\/li>\r\n \t<li>Use factoring by grouping to factor a trinomial<\/li>\r\n \t<li>Factor trinomials of the form\u00a0[latex]a{x}^{2}+bx+c[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the last section, we showed you how to factor polynomials with four terms by grouping.\u00a0 Trinomials of the form\u00a0[latex]a{x}^{2}+bx+c[\/latex] are slightly more complicated to factor. For 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\nBelow is a summary of the steps we will use followed by an example demonstrating how to use the step.\r\n<div class=\"textbox shaded\">\r\n<h3>Factoring Trinomials in the form [latex]ax^{2}+bx+c[\/latex]<\/h3>\r\nTo factor a trinomial in the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}r\\cdot{s}=a\\cdot{c}\\\\r+s=b\\end{array}[\/latex]<\/p>\r\nRewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex] and then use grouping and the distributive property to factor the polynomial.\r\n\r\n<\/div>\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]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\n&nbsp;\r\n\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+2x)+(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\">[latex]2x[\/latex] is the GCF of [latex](2x^2+2x)[\/latex] and [latex]3[\/latex] 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 as [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 does not matter.<\/p>\r\nNow let\u2019s see how this strategy works for factoring [latex]6z^{2}+11z+4[\/latex].\r\n\r\nIn this trinomial, [latex]a=6[\/latex], [latex]b=11[\/latex], and [latex]c=4[\/latex]. According to the strategy, you need to find two factors, <i>r<\/i> and <i>s<\/i>, whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since <i>ac<\/i> is positive and <i>b<\/i> is positive, you can be certain that the two factors you're looking for are also positive numbers.)\r\n<table style=\"width: 20%\">\r\n<thead>\r\n<tr>\r\n<th>Factors whose product is \u00a0[latex]24[\/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\\cdot24=24[\/latex]<\/td>\r\n<td>[latex]1+24=25[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2\\cdot12=24[\/latex]<\/td>\r\n<td>[latex]2+12=14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\\cdot8=24[\/latex]<\/td>\r\n<td>[latex]3+8=11[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4\\cdot6=24[\/latex]<\/td>\r\n<td>[latex]4+6=10[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere is only one combination where the product is [latex]24[\/latex] and the sum is [latex]11[\/latex], and that is when [latex]r=3[\/latex], and [latex]s=8[\/latex]. Let\u2019s use these values to factor the original trinomial.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]6z^{2}+11z+4[\/latex].\r\n[reveal-answer q=\"796129\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"796129\"]Rewrite the middle term, [latex]11z[\/latex], as [latex]3z + 8z[\/latex] (from the chart above.)\r\n<p style=\"text-align: center\">[latex]6z^{2}+3z+8z+4[\/latex]<\/p>\r\nGroup pairs. Use grouping to consider the terms in pairs.\r\n<p style=\"text-align: center\">[latex]\\left(6z^{2}+3z\\right)+\\left(8z+4\\right)[\/latex]<\/p>\r\nFactor [latex]3[\/latex]<i>z<\/i> out of the first group and [latex]4[\/latex] out of the second group.\r\n<p style=\"text-align: center\">[latex]3z\\left(2z+1\\right)+4\\left(2z+1\\right)[\/latex]<\/p>\r\nFactor out [latex]\\left(2z+1\\right)[\/latex].\r\n<p style=\"text-align: center\">[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow, let's look at an example where c is negative.\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 Solution[\/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>[latex]29[\/latex]<\/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>[latex]13[\/latex]<\/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>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]p=-3[\/latex] and [latex]q=10[\/latex].\r\n\r\n[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]\r\n\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[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n\r\n<span style=\"font-size: 1rem;text-align: initial\">We can summarize our process in the following way:<\/span>\r\n<div class=\"textbox\">\r\n<h3>How To: 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\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]115262[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"font-size: 1rem;text-align: initial\">In the following video, we present another example of factoring a trinomial using grouping. \u00a0In this example, the middle term, b, is negative. Note how having a negative middle term and a positive c term influence the options for r and s when factoring.<\/span>\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/agDaQ_cZnNc\r\n\r\nUsing this strategy, factoring trinomials becomes quick and kind of fun once you get the idea. \u00a0Give the next examples 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 Solution[\/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].\r\n\r\n[latex]9[\/latex] and [latex]18[\/latex] 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 for you to try where the 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 Solution[\/reveal-answer]\r\n[hidden-answer a=\"806715\"]\r\n\r\n&nbsp;\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\n<h3>Prime Trinomials<\/h3>\r\nBefore going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial [latex]2z^{2}+35z+7[\/latex], for instance. Can you think of two integers whose sum is [latex]b=35[\/latex] and whose product is [latex]a\\cdot{c}=2\\cdot7=14[\/latex]? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.\u00a0 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.\r\n<div class=\"bcc-box bcc-info\" style=\"text-align: left\">\r\n<h3>Example<\/h3>\r\nFactor [latex]7x^{2}-16x\u20135[\/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]7\\cdot{-5}=-35[\/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\nThe trinomial cannot be factored. None of the factors add up to [latex]-16[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Factoring out a GCF<\/h3>\r\nNow that you are familiar with the grouping method for factoring, we will now start to look at some strategies that are useful for making factoring easier.\u00a0 Your first step when factoring should always be to look for common factors for the three terms.\u00a0 Consider the examples below.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Trinomial<\/th>\r\n<th>Factor out Common Factor<\/th>\r\n<th>Factored<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]2x^{2}+10x+12[\/latex]<\/td>\r\n<td>[latex]2(x^{2}+5x+6)[\/latex]<\/td>\r\n<td>[latex]2\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22125a^{2}\u221215a\u221210[\/latex]<\/td>\r\n<td>[latex]\u22125(a^{2}+3a+2)[\/latex]<\/td>\r\n<td>[latex]\u22125\\left(a+2\\right)\\left(a+1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]c^{3}\u20138c^{2}+15c[\/latex]<\/td>\r\n<td>[latex]c\\left(c^{2}\u20138c+15\\right)[\/latex]<\/td>\r\n<td>[latex]c\\left(c\u20135\\right)\\left(c\u20133\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y^{4}\u20139y^{3}\u201310y^{2}[\/latex]<\/td>\r\n<td>[latex]y^{2}\\left(y^{2}\u20139y\u201310\\right)[\/latex]<\/td>\r\n<td>[latex]y^{2}\\left(y\u201310\\right)\\left(y+1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]3x^{3}\u20133x^{2}\u201390x[\/latex].\r\n\r\n[reveal-answer q=\"298928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"298928\"]Since 3 is a common factor for the three terms, factor out the 3.\r\n<p style=\"text-align: center\">[latex]3\\left(x^{3}\u2013x^{2}\u201330x\\right)[\/latex]<\/p>\r\n<i>x<\/i> is also a common factor, so factor out <i>x<\/i>.\r\n<p style=\"text-align: center\">[latex]3x\\left(x^{2}\u2013x\u201330\\right)[\/latex]<\/p>\r\nNow you can factor the trinomial\u00a0[latex]x^{2}\u2013x\u201330[\/latex]. To find <i>r<\/i> and <i>s<\/i>, identify two numbers whose product is [latex]\u221230[\/latex] and whose sum is [latex]\u22121[\/latex].\r\n\r\nThe pair of factors is [latex]\u22126[\/latex] and [latex]5[\/latex]. So replace [latex]\u2013x[\/latex] with [latex]\u22126x+5x[\/latex].\r\n<p style=\"text-align: center\">[latex]3x\\left(x^{2}\u20136x+5x\u201330\\right)[\/latex]<\/p>\r\nUse grouping to consider the terms in pairs.\r\n<p style=\"text-align: center\">[latex]3x\\left[\\left(x^{2}\u20136x\\right)+\\left(5x\u201330\\right)\\right][\/latex]<\/p>\r\nFactor <i>x<\/i> out of the first group and factor [latex]5[\/latex] out of the second group.\r\n<p style=\"text-align: center\">[latex]3x\\left[\\left(x\\left(x\u20136\\right)\\right)+5\\left(x\u20136\\right)\\right][\/latex]<\/p>\r\nThen factor out [latex]x\u20136[\/latex].\r\n<p style=\"text-align: center\">[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe general form of trinomials with a leading coefficient of <i>a<\/i> is [latex]ax^{2}+bx+c[\/latex]. Sometimes the factor of <i>a<\/i> can be factored as you saw above; this happens when <i>a<\/i> can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an [latex]x^{2}[\/latex]\u00a0term, instead of an [latex]ax^{2}[\/latex]\u00a0term.\r\n<h3>Factoring out a Negative<\/h3>\r\nIn some situations, <i>a<\/i> is negative, as in [latex]\u22124h^{2}+11h+3[\/latex]. It often makes sense to factor out [latex]\u22121[\/latex] as the first step in factoring, as doing so will change the sign of [latex]ax^{2}[\/latex]\u00a0from negative to positive, making the remaining trinomial easier to factor.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]\u22124h^{2}+11h+3[\/latex]\r\n[reveal-answer q=\"471034\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471034\"]Factor [latex]\u22121[\/latex] out of the trinomial. Notice that the signs of all three terms have changed.\r\n<p style=\"text-align: center\">[latex]\u22121\\left(4h^{2}\u201311h\u20133\\right)[\/latex]<\/p>\r\nTo factor the trinomial, you need to figure out how to rewrite [latex]\u221211h[\/latex].\u00a0The product of [latex]rs=4\\cdot\u22123=\u221212[\/latex], and the sum of [latex]rs=\u221211[\/latex].\r\n<table style=\"width: 20%\">\r\n<tbody>\r\n<tr>\r\n<td>[latex]r\\cdot{s}=\u221212[\/latex]<\/td>\r\n<td>[latex]r+s=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u221212\\cdot1=\u221212[\/latex]<\/td>\r\n<td>[latex]\u221212+1=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22126\\cdot2=\u221212[\/latex]<\/td>\r\n<td>[latex]\u22126+2=\u22124[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22124\\cdot3=\u221212[\/latex]<\/td>\r\n<td>[latex]\u22124+3=\u22121[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRewrite the middle term\u00a0[latex]\u221211h[\/latex] as [latex]\u221212h+1h[\/latex].\r\n<p style=\"text-align: center\">[latex]\u22121\\left(4h^{2}\u201312h+1h\u20133\\right)[\/latex]<\/p>\r\nGroup terms.\r\n<p style=\"text-align: center\">[latex]\u22121\\left[\\left(4h^{2}\u201312h\\right)+\\left(1h\u20133\\right)\\right][\/latex]<\/p>\r\nFactor out [latex]4[\/latex]<i>h<\/i> from the first pair. The second group cannot be factored further, but you can write it as [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0since [latex]+1\\left(h\u20133\\right)=\\left(h\u20133\\right)[\/latex]. This helps with factoring in the next step.\r\n<p style=\"text-align: center\">[latex]\u22121\\left[4h\\left(h\u20133\\right)+1\\left(h\u20133\\right)\\right][\/latex]<\/p>\r\nFactor out a common factor of [latex]\\left(h\u20133\\right)[\/latex]. Notice you are left with [latex]\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]; the [latex]+1[\/latex] comes from the term [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0in the previous step.\r\n<p style=\"text-align: center\">[latex]\u22121\\left[\\left(h\u20133\\right)\\left(4h+1\\right)\\right][\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\u22121\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that the answer above can also be written as [latex]\\left(\u2212h+3\\right)\\left(4h+1\\right)[\/latex] or [latex]\\left(h\u20133\\right)\\left(\u22124h\u20131\\right)[\/latex] if you multiply [latex]\u22121[\/latex]\u00a0times one of the other factors.\r\n\r\nIn the following video example, we will factor a trinomial whose leading term is negative.\u00a0 You will see how, by factoring out the negative sign, factoring the trinomial becomes easier.\r\n\r\nhttps:\/\/youtu.be\/zDAMjdBfkDs","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcome<\/h3>\n<ul>\n<li>Apply an algorithm to rewrite a trinomial as a four term polynomial and factor<\/li>\n<li>Use factoring by grouping to factor a trinomial<\/li>\n<li>Factor trinomials of the form\u00a0[latex]a{x}^{2}+bx+c[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p>In the last section, we showed you how to factor polynomials with four terms by grouping.\u00a0 Trinomials of the form\u00a0[latex]a{x}^{2}+bx+c[\/latex] are slightly more complicated to factor. For 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>Below is a summary of the steps we will use followed by an example demonstrating how to use the step.<\/p>\n<div class=\"textbox shaded\">\n<h3>Factoring Trinomials in the form [latex]ax^{2}+bx+c[\/latex]<\/h3>\n<p>To factor a trinomial in the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i><\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}r\\cdot{s}=a\\cdot{c}\\\\r+s=b\\end{array}[\/latex]<\/p>\n<p>Rewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex] and then use grouping and the distributive property to factor the polynomial.<\/p>\n<\/div>\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]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>&nbsp;<\/p>\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+2x)+(3x+3)[\/latex]<\/p>\n<p style=\"text-align: left\">Identify the GCF of each binomial.<\/p>\n<p style=\"text-align: left\">[latex]2x[\/latex] is the GCF of [latex](2x^2+2x)[\/latex] and [latex]3[\/latex] 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 as [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 does not matter.<\/p>\n<p>Now let\u2019s see how this strategy works for factoring [latex]6z^{2}+11z+4[\/latex].<\/p>\n<p>In this trinomial, [latex]a=6[\/latex], [latex]b=11[\/latex], and [latex]c=4[\/latex]. According to the strategy, you need to find two factors, <i>r<\/i> and <i>s<\/i>, whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since <i>ac<\/i> is positive and <i>b<\/i> is positive, you can be certain that the two factors you&#8217;re looking for are also positive numbers.)<\/p>\n<table style=\"width: 20%\">\n<thead>\n<tr>\n<th>Factors whose product is \u00a0[latex]24[\/latex]<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot24=24[\/latex]<\/td>\n<td>[latex]1+24=25[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2\\cdot12=24[\/latex]<\/td>\n<td>[latex]2+12=14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\\cdot8=24[\/latex]<\/td>\n<td>[latex]3+8=11[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4\\cdot6=24[\/latex]<\/td>\n<td>[latex]4+6=10[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>There is only one combination where the product is [latex]24[\/latex] and the sum is [latex]11[\/latex], and that is when [latex]r=3[\/latex], and [latex]s=8[\/latex]. Let\u2019s use these values to factor the original trinomial.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]6z^{2}+11z+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q796129\">Show Solution<\/span><\/p>\n<div id=\"q796129\" class=\"hidden-answer\" style=\"display: none\">Rewrite the middle term, [latex]11z[\/latex], as [latex]3z + 8z[\/latex] (from the chart above.)<\/p>\n<p style=\"text-align: center\">[latex]6z^{2}+3z+8z+4[\/latex]<\/p>\n<p>Group pairs. Use grouping to consider the terms in pairs.<\/p>\n<p style=\"text-align: center\">[latex]\\left(6z^{2}+3z\\right)+\\left(8z+4\\right)[\/latex]<\/p>\n<p>Factor [latex]3[\/latex]<i>z<\/i> out of the first group and [latex]4[\/latex] out of the second group.<\/p>\n<p style=\"text-align: center\">[latex]3z\\left(2z+1\\right)+4\\left(2z+1\\right)[\/latex]<\/p>\n<p>Factor out [latex]\\left(2z+1\\right)[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Now, let&#8217;s look at an example where c is negative.<\/p>\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 Solution<\/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>[latex]29[\/latex]<\/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>[latex]13[\/latex]<\/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>[latex]7[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So [latex]p=-3[\/latex] and [latex]q=10[\/latex].<\/p>\n<p>[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]<\/p>\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<\/div>\n<\/div>\n<div>\n<p><span style=\"font-size: 1rem;text-align: initial\">We can summarize our process in the following way:<\/span><\/p>\n<div class=\"textbox\">\n<h3>How To: 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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm115262\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115262&theme=oea&iframe_resize_id=ohm115262&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"font-size: 1rem;text-align: initial\">In the following video, we present another example of factoring a trinomial using grouping. \u00a0In this example, the middle term, b, is negative. Note how having a negative middle term and a positive c term influence the options for r and s when factoring.<\/span><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" 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>Using this strategy, factoring trinomials becomes quick and kind of fun once you get the idea. \u00a0Give the next examples 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 Solution<\/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].<\/p>\n<p>[latex]9[\/latex] and [latex]18[\/latex] 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 for you to try where the 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 Solution<\/span><\/p>\n<div id=\"q806715\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/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]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<h3>Prime Trinomials<\/h3>\n<p>Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial [latex]2z^{2}+35z+7[\/latex], for instance. Can you think of two integers whose sum is [latex]b=35[\/latex] and whose product is [latex]a\\cdot{c}=2\\cdot7=14[\/latex]? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.\u00a0 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\u20135[\/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]7\\cdot{-5}=-35[\/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<p>The trinomial cannot be factored. None of the factors add up to [latex]-16[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Factoring out a GCF<\/h3>\n<p>Now that you are familiar with the grouping method for factoring, we will now start to look at some strategies that are useful for making factoring easier.\u00a0 Your first step when factoring should always be to look for common factors for the three terms.\u00a0 Consider the examples below.<\/p>\n<table>\n<thead>\n<tr>\n<th>Trinomial<\/th>\n<th>Factor out Common Factor<\/th>\n<th>Factored<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]2x^{2}+10x+12[\/latex]<\/td>\n<td>[latex]2(x^{2}+5x+6)[\/latex]<\/td>\n<td>[latex]2\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22125a^{2}\u221215a\u221210[\/latex]<\/td>\n<td>[latex]\u22125(a^{2}+3a+2)[\/latex]<\/td>\n<td>[latex]\u22125\\left(a+2\\right)\\left(a+1\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]c^{3}\u20138c^{2}+15c[\/latex]<\/td>\n<td>[latex]c\\left(c^{2}\u20138c+15\\right)[\/latex]<\/td>\n<td>[latex]c\\left(c\u20135\\right)\\left(c\u20133\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]y^{4}\u20139y^{3}\u201310y^{2}[\/latex]<\/td>\n<td>[latex]y^{2}\\left(y^{2}\u20139y\u201310\\right)[\/latex]<\/td>\n<td>[latex]y^{2}\\left(y\u201310\\right)\\left(y+1\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]3x^{3}\u20133x^{2}\u201390x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q298928\">Show Solution<\/span><\/p>\n<div id=\"q298928\" class=\"hidden-answer\" style=\"display: none\">Since 3 is a common factor for the three terms, factor out the 3.<\/p>\n<p style=\"text-align: center\">[latex]3\\left(x^{3}\u2013x^{2}\u201330x\\right)[\/latex]<\/p>\n<p><i>x<\/i> is also a common factor, so factor out <i>x<\/i>.<\/p>\n<p style=\"text-align: center\">[latex]3x\\left(x^{2}\u2013x\u201330\\right)[\/latex]<\/p>\n<p>Now you can factor the trinomial\u00a0[latex]x^{2}\u2013x\u201330[\/latex]. To find <i>r<\/i> and <i>s<\/i>, identify two numbers whose product is [latex]\u221230[\/latex] and whose sum is [latex]\u22121[\/latex].<\/p>\n<p>The pair of factors is [latex]\u22126[\/latex] and [latex]5[\/latex]. So replace [latex]\u2013x[\/latex] with [latex]\u22126x+5x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]3x\\left(x^{2}\u20136x+5x\u201330\\right)[\/latex]<\/p>\n<p>Use grouping to consider the terms in pairs.<\/p>\n<p style=\"text-align: center\">[latex]3x\\left[\\left(x^{2}\u20136x\\right)+\\left(5x\u201330\\right)\\right][\/latex]<\/p>\n<p>Factor <i>x<\/i> out of the first group and factor [latex]5[\/latex] out of the second group.<\/p>\n<p style=\"text-align: center\">[latex]3x\\left[\\left(x\\left(x\u20136\\right)\\right)+5\\left(x\u20136\\right)\\right][\/latex]<\/p>\n<p>Then factor out [latex]x\u20136[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The general form of trinomials with a leading coefficient of <i>a<\/i> is [latex]ax^{2}+bx+c[\/latex]. Sometimes the factor of <i>a<\/i> can be factored as you saw above; this happens when <i>a<\/i> can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an [latex]x^{2}[\/latex]\u00a0term, instead of an [latex]ax^{2}[\/latex]\u00a0term.<\/p>\n<h3>Factoring out a Negative<\/h3>\n<p>In some situations, <i>a<\/i> is negative, as in [latex]\u22124h^{2}+11h+3[\/latex]. It often makes sense to factor out [latex]\u22121[\/latex] as the first step in factoring, as doing so will change the sign of [latex]ax^{2}[\/latex]\u00a0from negative to positive, making the remaining trinomial easier to factor.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]\u22124h^{2}+11h+3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q471034\">Show Solution<\/span><\/p>\n<div id=\"q471034\" class=\"hidden-answer\" style=\"display: none\">Factor [latex]\u22121[\/latex] out of the trinomial. Notice that the signs of all three terms have changed.<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left(4h^{2}\u201311h\u20133\\right)[\/latex]<\/p>\n<p>To factor the trinomial, you need to figure out how to rewrite [latex]\u221211h[\/latex].\u00a0The product of [latex]rs=4\\cdot\u22123=\u221212[\/latex], and the sum of [latex]rs=\u221211[\/latex].<\/p>\n<table style=\"width: 20%\">\n<tbody>\n<tr>\n<td>[latex]r\\cdot{s}=\u221212[\/latex]<\/td>\n<td>[latex]r+s=\u221211[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u221212\\cdot1=\u221212[\/latex]<\/td>\n<td>[latex]\u221212+1=\u221211[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22126\\cdot2=\u221212[\/latex]<\/td>\n<td>[latex]\u22126+2=\u22124[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22124\\cdot3=\u221212[\/latex]<\/td>\n<td>[latex]\u22124+3=\u22121[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Rewrite the middle term\u00a0[latex]\u221211h[\/latex] as [latex]\u221212h+1h[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left(4h^{2}\u201312h+1h\u20133\\right)[\/latex]<\/p>\n<p>Group terms.<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left[\\left(4h^{2}\u201312h\\right)+\\left(1h\u20133\\right)\\right][\/latex]<\/p>\n<p>Factor out [latex]4[\/latex]<i>h<\/i> from the first pair. The second group cannot be factored further, but you can write it as [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0since [latex]+1\\left(h\u20133\\right)=\\left(h\u20133\\right)[\/latex]. This helps with factoring in the next step.<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left[4h\\left(h\u20133\\right)+1\\left(h\u20133\\right)\\right][\/latex]<\/p>\n<p>Factor out a common factor of [latex]\\left(h\u20133\\right)[\/latex]. Notice you are left with [latex]\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]; the [latex]+1[\/latex] comes from the term [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0in the previous step.<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left[\\left(h\u20133\\right)\\left(4h+1\\right)\\right][\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\u22121\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Note that the answer above can also be written as [latex]\\left(\u2212h+3\\right)\\left(4h+1\\right)[\/latex] or [latex]\\left(h\u20133\\right)\\left(\u22124h\u20131\\right)[\/latex] if you multiply [latex]\u22121[\/latex]\u00a0times one of the other factors.<\/p>\n<p>In the following video example, we will factor a trinomial whose leading term is negative.\u00a0 You will see how, by factoring out the negative sign, factoring the trinomial becomes easier.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" 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\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-16323\">\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>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) for Lumen Learning. <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><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>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><\/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":169554,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/agDaQ_cZnNc\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Factor a Trinomial in the Form -ax^2+bx+c Using the Grouping Technique\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen 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