{"id":2723,"date":"2016-07-18T21:20:27","date_gmt":"2016-07-18T21:20:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2723"},"modified":"2019-08-06T18:56:57","modified_gmt":"2019-08-06T18:56:57","slug":"read-factor-by-grouping","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-factor-by-grouping\/","title":{"raw":"Factor by Grouping","rendered":"Factor by Grouping"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcome<\/h3>\r\n<ul>\r\n \t<li>Factor a trinomial with\u00a0leading coefficient \u2260\u00a0[latex]1[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nTrinomials with leading coefficients other than\u00a0[latex]1[\/latex] are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.\r\n\r\nThe first step in this process is to figure out what two numbers to use to re-write the\u00a0<em>x<\/em> term as the sum of two new terms. Making a table to keep track of your work is helpful. We are looking for two numbers with a product of [latex]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\n\r\n<div class=\"textbox\" style=\"text-align: center;\">\r\n<h3 style=\"text-align: left;\">\u00a0A General Note: Factor by Grouping<\/h3>\r\n<p style=\"text-align: left;\">To factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately. Then we factor out the GCF of the entire expression.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.\r\n[reveal-answer q=\"658545\"]Show 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<span style=\"font-size: 1rem; text-align: initial;\">In the following video, we present one more example of factoring a trinomial whose leading coefficient is not 1 using the grouping method.<\/span>\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/agDaQ_cZnNc\r\n\r\nFactoring trinomials whose leading coefficient is not\u00a0[latex]1[\/latex] becomes quick and kind of fun once you get the idea. \u00a0Give the next example a try on your own before you look at the solution.\r\n<div style=\"text-align: center;\">\r\n<p style=\"text-align: left;\">We will show two more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]2{x}^{2}+9x+9[\/latex].\r\n[reveal-answer q=\"834453\"]Show 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 where the x term is positive and c is negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]6{x}^{2}+x - 1[\/latex].\r\n[reveal-answer q=\"806715\"]Show 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\nIn the following video example, we will factor a trinomial whose leading term is negative.\r\n\r\nhttps:\/\/youtu.be\/zDAMjdBfkDs\r\n\r\nFor our last example, you will see that sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.\r\n<div class=\"bcc-box bcc-info\" style=\"text-align: left;\">\r\n<h3>Example<\/h3>\r\nFactor [latex]7x^{2}-16x\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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcome<\/h3>\n<ul>\n<li>Factor a trinomial with\u00a0leading coefficient \u2260\u00a0[latex]1[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p>Trinomials with leading coefficients other than\u00a0[latex]1[\/latex] are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.<\/p>\n<p>The first step in this process is to figure out what two numbers to use to re-write the\u00a0<em>x<\/em> term as the sum of two new terms. Making a table to keep track of your work is helpful. We are looking for two numbers with a product of [latex]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<div class=\"textbox\" style=\"text-align: center;\">\n<h3 style=\"text-align: left;\">\u00a0A General Note: Factor by Grouping<\/h3>\n<p style=\"text-align: left;\">To factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately. Then we factor out the GCF of the entire expression.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q658545\">Show 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<p><span style=\"font-size: 1rem; text-align: initial;\">In the following video, we present one more example of factoring a trinomial whose leading coefficient is not 1 using the grouping method.<\/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>Factoring trinomials whose leading coefficient is not\u00a0[latex]1[\/latex] becomes quick and kind of fun once you get the idea. \u00a0Give the next example a try on your own before you look at the solution.<\/p>\n<div style=\"text-align: center;\">\n<p style=\"text-align: left;\">We will show two more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]2{x}^{2}+9x+9[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834453\">Show 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 where the x term is positive and c is negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]6{x}^{2}+x - 1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q806715\">Show 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<p>In the following video example, we will factor a trinomial whose leading term is negative.<\/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>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\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-2723\">\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":21,"menu_order":5,"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 Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/zDAMjdBfkDs\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 12: Factoring, from Developmental Math: 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