{"id":2942,"date":"2024-02-09T20:30:31","date_gmt":"2024-02-09T20:30:31","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&#038;p=2942"},"modified":"2026-03-24T18:34:32","modified_gmt":"2026-03-24T18:34:32","slug":"9-6-4-factoring-trinomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/9-6-4-factoring-trinomials\/","title":{"raw":"9.6.4: Factoring Trinomials","rendered":"9.6.4: Factoring Trinomials"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h1>Learning Outcome<\/h1>\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 \t<li>Factoring perfect square trinomials<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h1>Key words<\/h1>\r\n<ul>\r\n \t<li><strong>Prime<\/strong>: having only two factors; itself and 1<\/li>\r\n \t<li><span style=\"font-size: 1rem; text-align: initial;\"><strong>Perfect square trinomial<\/strong>: any trinomial that factors into two identical binomials that can be written as the square of that binomial<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the last section, we looked at factoring trinomials of the form\u00a0[latex]a{x}^{2}+bx+c[\/latex]\u00a0<strong>by grouping.<\/strong>\u00a0We can use a similar method to factor trinomials of the form [latex]ax^2+bx+c[\/latex].\u00a0 The only difference is that we have to include the leading coefficient [latex]a[\/latex] in the process.\u00a0 We need to find to numbers [latex]r[\/latex] and [latex]s[\/latex] that are factors of the product [latex]ac[\/latex] and that sum to [latex]b[\/latex].\r\n\r\nThe 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 steps.\r\n<h3 class=\"textbox shaded\">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\nThe first step in this process is to figure out what two numbers to use to re-write the [latex]x[\/latex]-term as the sum of two new terms. Making a table to keep track of our work is helpful. We are looking for two numbers with a product of [latex]ac=6[\/latex] and a sum of [latex]b=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]ac=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]r=2 \\text{ and }s=3[\/latex] will give the correct\u00a0[latex]x[\/latex]-coefficient of [latex]5[\/latex], 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]. We use these 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;\">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 we factor out the GCF. This is purely a matter of preference. Multiplication is commutative, so order does not matter.<\/p>\r\n&nbsp;\r\n\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, we need to find two factors, [latex]r[\/latex] and\u00a0[latex]s[\/latex], whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. We can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since\u00a0[latex]ac[\/latex]\u00a0is positive and\u00a0[latex]b[\/latex] is positive, we can be certain that the two factors we'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]ac=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<h4 class=\"bcc-box bcc-info\">Example<\/h4>\r\nFactor [latex]6z^{2}+11z+4[\/latex].\r\n<h4 class=\"bcc-box bcc-info\">Solution<\/h4>\r\nRewrite 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]3z[\/latex] 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 class=\"bcc-box bcc-info\">Answer<\/h4>\r\n[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]10y^2+23y+12[\/latex]\r\n\r\n[reveal-answer q=\"hjm395\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm395\"][latex](5y+4)(2y+3)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nNow, let's look at an example where [latex]c[\/latex] 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<h4>Solution<\/h4>\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]r=-3[\/latex] and [latex]s=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}+rx+sx+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<h4>Answer<\/h4>\r\n[latex]5{x}^{2}+7x - 6=\\left(5x - 3\\right)\\left(x+2\\right)[\/latex]\r\n\r\n&nbsp;\r\n\r\nWe can check our work by multiplying to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].\r\n\r\n<\/div>\r\n<div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]4x^2-8x-5[\/latex]\r\n\r\n[reveal-answer q=\"hjm779\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm779\"][latex](2x-5)(2x+1)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\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 shaded\">\r\n<h3>factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex]<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]ac[\/latex].<\/li>\r\n \t<li>Find [latex]r[\/latex] and [latex]s[\/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}+rx+sx+c[\/latex].<\/li>\r\n \t<li>Pull out the GCF of [latex]a{x}^{2}+rx[\/latex].<\/li>\r\n \t<li>Pull out the GCF of [latex]sx+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;\">The following video presents another example of factoring a trinomial using grouping. \u00a0In this example, the [latex]x[\/latex]-term, [latex]b[\/latex], is negative. Note how having a negative middle term and a positive constant term influences the options for r and s when factoring.<\/span>\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/agDaQ_cZnNc\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.6.4-1.docx\">Transcript-9.6.4-1<\/a>\r\n<div style=\"text-align: center;\"><\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]2{x}^{2}+9x+9[\/latex].\r\n<h4>Solution<\/h4>\r\nFind two numbers [latex]r[\/latex] and [latex]s[\/latex] such that [latex]r\\cdot{s}=18[\/latex] and [latex]r + s = 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\r\n<h4>Answer<\/h4>\r\n[latex]2{x}^{2}+9x+9=(x+3)(2x+3)[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]12x^2+25x+7[\/latex]\r\n\r\n[reveal-answer q=\"hjm875\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm875\"][latex](4x+7)(3x+1)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nHere is an\u00a0example where the constant term is negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]6{x}^{2}+x - 1[\/latex].\r\n<h4>Solution<\/h4>\r\nDetermine [latex]ac=6(-1)=-6[\/latex]. We need two factors of [latex]ac[\/latex] that sum to [latex]b=1[\/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]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\r\n<h4>Answer<\/h4>\r\n[latex]6{x}^{2}+x - 1=(2x+1)(3x-1)[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]6x^2+53x-9[\/latex]\r\n\r\n[reveal-answer q=\"hjm323\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm323\"][latex](6x-1)(x+9)[\/latex][\/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. Indeed, there are more trinomials that can't be factored than trinomials that can be factored. Take the trinomial [latex]2z^{2}+35z+7[\/latex], for instance. Are there 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 <em><strong>prime<\/strong><\/em> trinomial.\u00a0 We will sometimes encounter polynomials that, despite our 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<h4>Solution<\/h4>\r\nFind [latex]r, s[\/latex] such that [latex]r\\cdot{s}=-35\\text{ and }r+s=-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\nWe have exhausted all integer pairs, so the trinomial cannot be factored. None of the factors add up to [latex]-16[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]7x^{2}-16x\u20135[\/latex] is prime.\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]5x^2-3x+1[\/latex]\r\n\r\n[reveal-answer q=\"hjm277\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm277\"][latex]5x^2-3x+1[\/latex] is prime[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Factoring out a GCF<\/h3>\r\nConsider\u00a0[latex]2x^{2}+10x+12[\/latex]. Since [latex]ac=2\\cdot 12=24[\/latex] and [latex]b=10[\/latex], the values [latex]r=6[\/latex] and [latex]s=4[\/latex] can be used to rewrite the trinomial as [latex]2x^2+6x+4x+12[\/latex]. Factoring each pair leads to [latex]2x(x+3)+4(x+3)=(2x+4)(x+3)[\/latex].\u00a0 But notice that this is not completely factored because the binomial [latex]2x+4[\/latex] can be further factored by pulling out the greatest common factor of [latex]2[\/latex] resulting in [latex]2(x+2)[\/latex]. This means that\u00a0[latex]2x^{2}+10x+12[\/latex] factors to [latex]2(x+2)(x+3)[\/latex].\r\n\r\nWe really should have noticed the GCF right away:\u00a0[latex]2x^{2}+10x+12=2(x^2+5x+6)[\/latex]. Pulling the GCF out first makes the rest of the factoring easier as we are dealing with smaller values of [latex]a, b[\/latex] and [latex]c[\/latex].\u00a0 In this case,\r\n<p style=\"text-align: center;\">[latex]\\;\\;\\;\\;2x^{2}+10x+12\\\\=2(x^2+5x+6)\\\\=2(x+3)(x+2)[\/latex]<\/p>\r\nExactly the same answer we got before.\r\n\r\nConsequently, we should always start factoring by looking for a GCF for all terms in the trinomial. Once we have identified and pulled out the greatest common factor, we can factor the remaining trinomial as usual.\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<h4>Solution<\/h4>\r\nSince [latex]3x[\/latex] is a common factor for the three terms, factor out the[latex]3x[\/latex] first:\r\n<p style=\"text-align: center;\">[latex]3x\\left(x^{2}\u2013x\u201330\\right)[\/latex]<\/p>\r\nNow we 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]ac=\u221230[\/latex] and whose sum is [latex]b=\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 [latex]x[\/latex] 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^{3}\u20133x^{2}\u201390x=3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]30x^4+27x^3-27x^2[\/latex]\r\n\r\n[reveal-answer q=\"hjm432\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm432\"][latex]3x^2(5x-3)(2x+3)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Factoring out a Negative<\/h3>\r\nIn some situations, [latex]a[\/latex] 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<h4>Solution<\/h4>\r\nFactor [latex]\u22121[\/latex] out of the trinomial. Notice that the signs of all three terms will change to their opposites:\r\n<p style=\"text-align: center;\">[latex]\u22124h^{2}+11h+3[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]=\u22121\\left(4h^{2}\u201311h\u20133\\right)[\/latex]<\/p>\r\nTo factor the trinomial, we need to figure out how to rewrite [latex]\u221211h[\/latex].\u00a0The product [latex]rs=4\\cdot\u22123=\u221212[\/latex], and the sum [latex]r+s=\u221211[\/latex].\r\n<table style=\"width: 20%;\">\r\n<tbody>\r\n<tr>\r\n<th>The product [latex]rs=4\\cdot\u22123=\u221212[\/latex]<\/th>\r\n<th>The sum [latex]r+s=\u221211[\/latex]<\/th>\r\n<\/tr>\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 only common factor of the second group is [latex]1[\/latex], so we 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 we 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]\u22124h^{2}+11h+3=\u22121\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]\r\n\r\n<\/div>\r\nThe following video shows an example where the leading term is negative.\u00a0 We will see how, by factoring out the negative sign, factoring the trinomial becomes easier.\r\n\r\nhttps:\/\/youtu.be\/zDAMjdBfkDs\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.6.4-2.docx\">Transcript-9.6.4-2<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]-60x^5-35x^4+60x^3[\/latex]\r\n\r\n[reveal-answer q=\"hjm531\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm531\"][latex]-5x^3(4x-3)(3x+4)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]-9x^3+24x^2-16x[\/latex]\r\n\r\n[reveal-answer q=\"hjm722\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm722\"][latex]-x(3x-4)(3x-4)=-x{(3x-4)}^{2}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that [latex]-9x^3+24x^2-16x[\/latex] factors into\u00a0[latex]-x(3x-4)(3x-4)[\/latex], which can be written as\u00a0[latex]-x(3x-4)^2[\/latex].\r\n\r\n[latex](3x-4)^2[\/latex] is the square of the binomial\u00a0[latex](3x-4)[\/latex]. When\u00a0[latex](3x-4)[\/latex] is squared we get\u00a0[latex]{(3x-4)}^{2}=(3x-4)(3x-4)=9x^2-24x+16[\/latex]. The trinomial\u00a0[latex]9x^2-24x+16[\/latex] is known as a\u00a0<em><strong>perfect square trinomial<\/strong><\/em>.\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nFactor [latex]3x^4-18x^3+27x^2[\/latex]\r\n<h4>Solution<\/h4>\r\nLook for a GCF first: [latex]3x^2[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\;\\;\\;\\;3x^4-18x^3+27x^2[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]=3x^2(x^2-6x+9)[\/latex]<\/p>\r\n[latex]ac=9[\/latex] and [latex]b=-6[\/latex]:\r\n<table style=\"border-collapse: collapse; width: 7.98995%; height: 135px;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 23.0358%; text-align: center;\">[latex]ac=9[\/latex]<\/th>\r\n<th style=\"width: 60.7142%; text-align: center;\">Sum = [latex]6[\/latex]<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">9, 1<\/td>\r\n<td style=\"width: 50%; text-align: center;\">10<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">-9, -1<\/td>\r\n<td style=\"width: 50%; text-align: center;\">\u00a0-10<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">-9, 1<\/td>\r\n<td style=\"width: 50%; text-align: center;\">-8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">9, -1<\/td>\r\n<td style=\"width: 50%; text-align: center;\">8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">3, 3<\/td>\r\n<td style=\"width: 50%; text-align: center;\">6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">-3, -3<\/td>\r\n<td style=\"width: 50%; text-align: center;\">-6<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRewrite [latex]-6x[\/latex] as [latex]-3x-3x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]3x^2(x^2-3x-3x+9)[\/latex]<\/p>\r\nGroup and pull out the common factors:\r\n<p style=\"text-align: center;\">[latex]3x^2\\left [ x(x-3)-3(x-3)\\right ] [\/latex]<\/p>\r\nPull out the common factor [latex](x-3)[\/latex]:\r\n<p style=\"text-align: center;\">[latex]3x^2(x-3)(x-3)[\/latex]<\/p>\r\nRewrite the binomials as a perfect square:\r\n<p style=\"text-align: center;\">[latex]3x^2{(x-3)}^{2}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]3x^4-18x^3+27x^2=3x^2(x-3)^2[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]-25x^2+20x-4[\/latex]\r\n\r\n[reveal-answer q=\"hjm301\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm301\"][latex]-{(5x-2)}^{2}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h1>Learning Outcome<\/h1>\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<li>Factoring perfect square trinomials<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h1>Key words<\/h1>\n<ul>\n<li><strong>Prime<\/strong>: having only two factors; itself and 1<\/li>\n<li><span style=\"font-size: 1rem; text-align: initial;\"><strong>Perfect square trinomial<\/strong>: any trinomial that factors into two identical binomials that can be written as the square of that binomial<\/span><\/li>\n<\/ul>\n<\/div>\n<p>In the last section, we looked at factoring trinomials of the form\u00a0[latex]a{x}^{2}+bx+c[\/latex]\u00a0<strong>by grouping.<\/strong>\u00a0We can use a similar method to factor trinomials of the form [latex]ax^2+bx+c[\/latex].\u00a0 The only difference is that we have to include the leading coefficient [latex]a[\/latex] in the process.\u00a0 We need to find to numbers [latex]r[\/latex] and [latex]s[\/latex] that are factors of the product [latex]ac[\/latex] and that sum to [latex]b[\/latex].<\/p>\n<p>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 steps.<\/p>\n<h3 class=\"textbox shaded\">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<p>The first step in this process is to figure out what two numbers to use to re-write the [latex]x[\/latex]-term as the sum of two new terms. Making a table to keep track of our work is helpful. We are looking for two numbers with a product of [latex]ac=6[\/latex] and a sum of [latex]b=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]ac=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]r=2 \\text{ and }s=3[\/latex] will give the correct\u00a0[latex]x[\/latex]-coefficient of [latex]5[\/latex], 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]. We use these 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;\">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 we factor out the GCF. This is purely a matter of preference. Multiplication is commutative, so order does not matter.<\/p>\n<p>&nbsp;<\/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, we need to find two factors, [latex]r[\/latex] and\u00a0[latex]s[\/latex], whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. We can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since\u00a0[latex]ac[\/latex]\u00a0is positive and\u00a0[latex]b[\/latex] is positive, we can be certain that the two factors we&#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]ac=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<h4 class=\"bcc-box bcc-info\">Example<\/h4>\n<p>Factor [latex]6z^{2}+11z+4[\/latex].<\/p>\n<h4 class=\"bcc-box bcc-info\">Solution<\/h4>\n<p>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]3z[\/latex] 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 class=\"bcc-box bcc-info\">Answer<\/h4>\n<p>[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]10y^2+23y+12[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm395\">Show Answer<\/span><\/p>\n<div id=\"qhjm395\" class=\"hidden-answer\" style=\"display: none\">[latex](5y+4)(2y+3)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Now, let&#8217;s look at an example where [latex]c[\/latex] 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<h4>Solution<\/h4>\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]r=-3[\/latex] and [latex]s=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}+rx+sx+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<h4>Answer<\/h4>\n<p>[latex]5{x}^{2}+7x - 6=\\left(5x - 3\\right)\\left(x+2\\right)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>We can check our work by multiplying 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 class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]4x^2-8x-5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm779\">Show Answer<\/span><\/p>\n<div id=\"qhjm779\" class=\"hidden-answer\" style=\"display: none\">[latex](2x-5)(2x+1)[\/latex]<\/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 shaded\">\n<h3>factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex]<\/h3>\n<ol>\n<li>List factors of [latex]ac[\/latex].<\/li>\n<li>Find [latex]r[\/latex] and [latex]s[\/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}+rx+sx+c[\/latex].<\/li>\n<li>Pull out the GCF of [latex]a{x}^{2}+rx[\/latex].<\/li>\n<li>Pull out the GCF of [latex]sx+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;\">The following video presents another example of factoring a trinomial using grouping. \u00a0In this example, the [latex]x[\/latex]-term, [latex]b[\/latex], is negative. Note how having a negative middle term and a positive constant term influences 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><a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.6.4-1.docx\">Transcript-9.6.4-1<\/a><\/p>\n<div style=\"text-align: center;\"><\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]2{x}^{2}+9x+9[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Find two numbers [latex]r[\/latex] and [latex]s[\/latex] such that [latex]r\\cdot{s}=18[\/latex] and [latex]r + s = 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<h4>Answer<\/h4>\n<p>[latex]2{x}^{2}+9x+9=(x+3)(2x+3)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]12x^2+25x+7[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm875\">Show Answer<\/span><\/p>\n<div id=\"qhjm875\" class=\"hidden-answer\" style=\"display: none\">[latex](4x+7)(3x+1)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Here is an\u00a0example where the constant term is negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]6{x}^{2}+x - 1[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Determine [latex]ac=6(-1)=-6[\/latex]. We need two factors of [latex]ac[\/latex] that sum to [latex]b=1[\/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]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<h4>Answer<\/h4>\n<p>[latex]6{x}^{2}+x - 1=(2x+1)(3x-1)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]6x^2+53x-9[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm323\">Show Answer<\/span><\/p>\n<div id=\"qhjm323\" class=\"hidden-answer\" style=\"display: none\">[latex](6x-1)(x+9)[\/latex]<\/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. Indeed, there are more trinomials that can&#8217;t be factored than trinomials that can be factored. Take the trinomial [latex]2z^{2}+35z+7[\/latex], for instance. Are there 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 <em><strong>prime<\/strong><\/em> trinomial.\u00a0 We will sometimes encounter polynomials that, despite our 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<h4>Solution<\/h4>\n<p>Find [latex]r, s[\/latex] such that [latex]r\\cdot{s}=-35\\text{ and }r+s=-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>We have exhausted all integer pairs, so the trinomial cannot be factored. None of the factors add up to [latex]-16[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]7x^{2}-16x\u20135[\/latex] is prime.<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]5x^2-3x+1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm277\">Show Answer<\/span><\/p>\n<div id=\"qhjm277\" class=\"hidden-answer\" style=\"display: none\">[latex]5x^2-3x+1[\/latex] is prime<\/div>\n<\/div>\n<\/div>\n<h3>Factoring out a GCF<\/h3>\n<p>Consider\u00a0[latex]2x^{2}+10x+12[\/latex]. Since [latex]ac=2\\cdot 12=24[\/latex] and [latex]b=10[\/latex], the values [latex]r=6[\/latex] and [latex]s=4[\/latex] can be used to rewrite the trinomial as [latex]2x^2+6x+4x+12[\/latex]. Factoring each pair leads to [latex]2x(x+3)+4(x+3)=(2x+4)(x+3)[\/latex].\u00a0 But notice that this is not completely factored because the binomial [latex]2x+4[\/latex] can be further factored by pulling out the greatest common factor of [latex]2[\/latex] resulting in [latex]2(x+2)[\/latex]. This means that\u00a0[latex]2x^{2}+10x+12[\/latex] factors to [latex]2(x+2)(x+3)[\/latex].<\/p>\n<p>We really should have noticed the GCF right away:\u00a0[latex]2x^{2}+10x+12=2(x^2+5x+6)[\/latex]. Pulling the GCF out first makes the rest of the factoring easier as we are dealing with smaller values of [latex]a, b[\/latex] and [latex]c[\/latex].\u00a0 In this case,<\/p>\n<p style=\"text-align: center;\">[latex]\\;\\;\\;\\;2x^{2}+10x+12\\\\=2(x^2+5x+6)\\\\=2(x+3)(x+2)[\/latex]<\/p>\n<p>Exactly the same answer we got before.<\/p>\n<p>Consequently, we should always start factoring by looking for a GCF for all terms in the trinomial. Once we have identified and pulled out the greatest common factor, we can factor the remaining trinomial as usual.<\/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<h4>Solution<\/h4>\n<p>Since [latex]3x[\/latex] is a common factor for the three terms, factor out the[latex]3x[\/latex] first:<\/p>\n<p style=\"text-align: center;\">[latex]3x\\left(x^{2}\u2013x\u201330\\right)[\/latex]<\/p>\n<p>Now we 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]ac=\u221230[\/latex] and whose sum is [latex]b=\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 [latex]x[\/latex] 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^{3}\u20133x^{2}\u201390x=3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]30x^4+27x^3-27x^2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm432\">Show Answer<\/span><\/p>\n<div id=\"qhjm432\" class=\"hidden-answer\" style=\"display: none\">[latex]3x^2(5x-3)(2x+3)[\/latex]<\/div>\n<\/div>\n<\/div>\n<h3>Factoring out a Negative<\/h3>\n<p>In some situations, [latex]a[\/latex] 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<h4>Solution<\/h4>\n<p>Factor [latex]\u22121[\/latex] out of the trinomial. Notice that the signs of all three terms will change to their opposites:<\/p>\n<p style=\"text-align: center;\">[latex]\u22124h^{2}+11h+3[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]=\u22121\\left(4h^{2}\u201311h\u20133\\right)[\/latex]<\/p>\n<p>To factor the trinomial, we need to figure out how to rewrite [latex]\u221211h[\/latex].\u00a0The product [latex]rs=4\\cdot\u22123=\u221212[\/latex], and the sum [latex]r+s=\u221211[\/latex].<\/p>\n<table style=\"width: 20%;\">\n<tbody>\n<tr>\n<th>The product [latex]rs=4\\cdot\u22123=\u221212[\/latex]<\/th>\n<th>The sum [latex]r+s=\u221211[\/latex]<\/th>\n<\/tr>\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 only common factor of the second group is [latex]1[\/latex], so we 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 we 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]\u22124h^{2}+11h+3=\u22121\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]<\/p>\n<\/div>\n<p>The following video shows an example where the leading term is negative.\u00a0 We 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<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-content\/uploads\/sites\/5676\/2024\/02\/Transcript-9.6.4-2.docx\">Transcript-9.6.4-2<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]-60x^5-35x^4+60x^3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm531\">Show Answer<\/span><\/p>\n<div id=\"qhjm531\" class=\"hidden-answer\" style=\"display: none\">[latex]-5x^3(4x-3)(3x+4)[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]-9x^3+24x^2-16x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm722\">Show Answer<\/span><\/p>\n<div id=\"qhjm722\" class=\"hidden-answer\" style=\"display: none\">[latex]-x(3x-4)(3x-4)=-x{(3x-4)}^{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Notice that [latex]-9x^3+24x^2-16x[\/latex] factors into\u00a0[latex]-x(3x-4)(3x-4)[\/latex], which can be written as\u00a0[latex]-x(3x-4)^2[\/latex].<\/p>\n<p>[latex](3x-4)^2[\/latex] is the square of the binomial\u00a0[latex](3x-4)[\/latex]. When\u00a0[latex](3x-4)[\/latex] is squared we get\u00a0[latex]{(3x-4)}^{2}=(3x-4)(3x-4)=9x^2-24x+16[\/latex]. The trinomial\u00a0[latex]9x^2-24x+16[\/latex] is known as a\u00a0<em><strong>perfect square trinomial<\/strong><\/em>.<\/p>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Factor [latex]3x^4-18x^3+27x^2[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Look for a GCF first: [latex]3x^2[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\;\\;\\;\\;3x^4-18x^3+27x^2[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]=3x^2(x^2-6x+9)[\/latex]<\/p>\n<p>[latex]ac=9[\/latex] and [latex]b=-6[\/latex]:<\/p>\n<table style=\"border-collapse: collapse; width: 7.98995%; height: 135px;\">\n<tbody>\n<tr>\n<th style=\"width: 23.0358%; text-align: center;\">[latex]ac=9[\/latex]<\/th>\n<th style=\"width: 60.7142%; text-align: center;\">Sum = [latex]6[\/latex]<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\">9, 1<\/td>\n<td style=\"width: 50%; text-align: center;\">10<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\">-9, -1<\/td>\n<td style=\"width: 50%; text-align: center;\">\u00a0-10<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\">-9, 1<\/td>\n<td style=\"width: 50%; text-align: center;\">-8<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\">9, -1<\/td>\n<td style=\"width: 50%; text-align: center;\">8<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\">3, 3<\/td>\n<td style=\"width: 50%; text-align: center;\">6<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\">-3, -3<\/td>\n<td style=\"width: 50%; text-align: center;\">-6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Rewrite [latex]-6x[\/latex] as [latex]-3x-3x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]3x^2(x^2-3x-3x+9)[\/latex]<\/p>\n<p>Group and pull out the common factors:<\/p>\n<p style=\"text-align: center;\">[latex]3x^2\\left [ x(x-3)-3(x-3)\\right ][\/latex]<\/p>\n<p>Pull out the common factor [latex](x-3)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]3x^2(x-3)(x-3)[\/latex]<\/p>\n<p>Rewrite the binomials as a perfect square:<\/p>\n<p style=\"text-align: center;\">[latex]3x^2{(x-3)}^{2}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]3x^4-18x^3+27x^2=3x^2(x-3)^2[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]-25x^2+20x-4[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm301\">Show Answer<\/span><\/p>\n<div id=\"qhjm301\" class=\"hidden-answer\" style=\"display: none\">[latex]-{(5x-2)}^{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n","protected":false},"author":370291,"menu_order":12,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2942","chapter","type-chapter","status-publish","hentry"],"part":2917,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2942","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/users\/370291"}],"version-history":[{"count":3,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2942\/revisions"}],"predecessor-version":[{"id":3192,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2942\/revisions\/3192"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/parts\/2917"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2942\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/media?parent=2942"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2942"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/contributor?post=2942"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/license?post=2942"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}