{"id":1237,"date":"2016-10-21T21:10:03","date_gmt":"2016-10-21T21:10:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1237"},"modified":"2017-04-10T19:51:45","modified_gmt":"2017-04-10T19:51:45","slug":"factoring-basics","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/factoring-basics\/","title":{"raw":"Factoring Basics","rendered":"Factoring Basics"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Identify and factor the GCF of a polynomial<\/li>\r\n \t<li>Factor a Trinomial with Leading Coefficient 1<\/li>\r\n \t<li>Factor by Grouping<\/li>\r\n<\/ul>\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\nWhen we study fractions, we learn that the <strong>greatest common factor<\/strong> (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, [latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest number that divides evenly into both [latex]16[\/latex] and [latex]20[\/latex] The GCF of polynomials works the same way: [latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest polynomial that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].\r\n\r\nWhen factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Greatest Common Factor<\/h3>\r\nThe <strong>greatest common factor<\/strong> (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a polynomial expression, factor out the greatest common factor.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Identify the GCF of the coefficients.<\/li>\r\n \t<li>Identify the GCF of the variables.<\/li>\r\n \t<li>Combine to find the GCF of the expression.<\/li>\r\n \t<li>Determine what the GCF needs to be multiplied by to obtain each term in the expression.<\/li>\r\n \t<li>Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring the Greatest Common Factor<\/h3>\r\nFactor [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].\r\n\r\n[reveal-answer q=\"113189\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"113189\"]\r\nFirst, find the GCF of the expression. The GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. The GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions in the form [latex]{x}^{n}[\/latex] will always be the exponent of lowest degree.) And the GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. Combine these to find the GCF of the polynomial, [latex]3xy[\/latex].\r\n\r\nNext, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that [latex]3xy\\left(2{x}^{2}{y}^{2}\\right)=6{x}^{3}{y}^{3},3xy\\left(15xy\\right)=45{x}^{2}{y}^{2}[\/latex], and [latex]3xy\\left(7\\right)=21xy[\/latex].\r\n\r\nFinally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.\r\n<div style=\"text-align: center;\">[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]<\/div>\r\n<div>\r\n<h4>Analysis of the Solution<\/h4>\r\nAfter factoring, we can check our work by multiplying. Use the distributive property to confirm that [latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].\r\n\r\n<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div><\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]x\\left({b}^{2}-a\\right)+6\\left({b}^{2}-a\\right)[\/latex] by pulling out the GCF.\r\n\r\n[reveal-answer q=\"94532\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"94532\"][latex]\\left({b}^{2}-a\\right)\\left(x+6\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7886&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\nWatch this video to see more examples of how to factor the GCF from a trinomial.\r\n\r\nhttps:\/\/youtu.be\/3f1RFTIw2Ng\r\n<h2>Factoring a Trinomial with Leading Coefficient 1<\/h2>\r\nAlthough we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].\r\n\r\nTrinomials of the form [latex]{x}^{2}+bx+c[\/latex] can be factored by finding two numbers with a product of [latex]c[\/latex] and a sum of [latex]b[\/latex]. The trinomial [latex]{x}^{2}+10x+16[\/latex], for example, can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex] because the product of those numbers is [latex]16[\/latex] and their sum is [latex]10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nA trinomial of the form [latex]{x}^{2}+bx+c[\/latex] can be written in factored form as [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex] where [latex]pq=c[\/latex] and [latex]p+q=b[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h3>Can every trinomial be factored as a product of binomials?<\/h3>\r\n<em>No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it.<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]c[\/latex].<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nFactor [latex]{x}^{2}+2x - 15[\/latex].\r\n\r\n[reveal-answer q=\"88306\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"88306\"]\r\nWe have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]-15[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,-15[\/latex]<\/td>\r\n<td>[latex]-14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,15[\/latex]<\/td>\r\n<td>14<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-5[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,5[\/latex]<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can check our work by multiplying. Use FOIL to confirm that [latex]\\left(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div><\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h3>Does the order of the factors matter?<\/h3>\r\n<em>No. Multiplication is commutative, so the order of the factors does not matter.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]{x}^{2}-7x+6[\/latex].\r\n\r\n[reveal-answer q=\"823058\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"823058\"][latex]\\left(x - 6\\right)\\left(x - 1\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7897&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Factoring by Grouping<\/h2>\r\nTrinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Factor by Grouping<\/h3>\r\nTo factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.\r\n\r\n<\/div>\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 exercises\">\r\n<h3>Example: Factoring a Trinomial by Grouping<\/h3>\r\nFactor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.\r\n\r\n[reveal-answer q=\"806328\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"806328\"]\r\nWe have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.\r\n<table summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]-30[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,-30[\/latex]<\/td>\r\n<td>[latex]-29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1,30[\/latex]<\/td>\r\n<td>29<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2,-15[\/latex]<\/td>\r\n<td>[latex]-13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,15[\/latex]<\/td>\r\n<td>13<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3,-10[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3,10[\/latex]<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]p=-3[\/latex] and [latex]q=10[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill &amp; \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill &amp; \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill &amp; \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].\r\n\r\n<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div><\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor the following.\r\n<ol>\r\n \t<li>[latex]2{x}^{2}+9x+9[\/latex]<\/li>\r\n \t<li>[latex]6{x}^{2}+x - 1[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"343485\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"343485\"]\r\n<ol>\r\n \t<li>[latex]\\left(2x+3\\right)\\left(x+3\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\left(3x - 1\\right)\\left(2x+1\\right)[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7908&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\nIn the next video we show another example of how to factor a trinomial by grouping.\r\nhttps:\/\/youtu.be\/agDaQ_cZnNc","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Identify and factor the GCF of a polynomial<\/li>\n<li>Factor a Trinomial with Leading Coefficient 1<\/li>\n<li>Factor by Grouping<\/li>\n<\/ul>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><br \/>\nWhen we study fractions, we learn that the <strong>greatest common factor<\/strong> (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, [latex]4[\/latex] is the GCF of [latex]16[\/latex] and [latex]20[\/latex] because it is the largest number that divides evenly into both [latex]16[\/latex] and [latex]20[\/latex] The GCF of polynomials works the same way: [latex]4x[\/latex] is the GCF of [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex] because it is the largest polynomial that divides evenly into both [latex]16x[\/latex] and [latex]20{x}^{2}[\/latex].<\/p>\n<p>When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Greatest Common Factor<\/h3>\n<p>The <strong>greatest common factor<\/strong> (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a polynomial expression, factor out the greatest common factor.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Identify the GCF of the coefficients.<\/li>\n<li>Identify the GCF of the variables.<\/li>\n<li>Combine to find the GCF of the expression.<\/li>\n<li>Determine what the GCF needs to be multiplied by to obtain each term in the expression.<\/li>\n<li>Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring the Greatest Common Factor<\/h3>\n<p>Factor [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q113189\">Solution<\/span><\/p>\n<div id=\"q113189\" class=\"hidden-answer\" style=\"display: none\">\nFirst, find the GCF of the expression. The GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. The GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions in the form [latex]{x}^{n}[\/latex] will always be the exponent of lowest degree.) And the GCF of [latex]{y}^{3},{y}^{2}[\/latex], and [latex]y[\/latex] is [latex]y[\/latex]. Combine these to find the GCF of the polynomial, [latex]3xy[\/latex].<\/p>\n<p>Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that [latex]3xy\\left(2{x}^{2}{y}^{2}\\right)=6{x}^{3}{y}^{3},3xy\\left(15xy\\right)=45{x}^{2}{y}^{2}[\/latex], and [latex]3xy\\left(7\\right)=21xy[\/latex].<\/p>\n<p>Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.<\/p>\n<div style=\"text-align: center;\">[latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)[\/latex]<\/div>\n<div>\n<h4>Analysis of the Solution<\/h4>\n<p>After factoring, we can check our work by multiplying. Use the distributive property to confirm that [latex]\\left(3xy\\right)\\left(2{x}^{2}{y}^{2}+15xy+7\\right)=6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].<\/p>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor [latex]x\\left({b}^{2}-a\\right)+6\\left({b}^{2}-a\\right)[\/latex] by pulling out the GCF.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q94532\">Solution<\/span><\/p>\n<div id=\"q94532\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left({b}^{2}-a\\right)\\left(x+6\\right)[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7886&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>Watch this video to see more examples of how to factor the GCF from a trinomial.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 2:  Identify GCF and Factor a Trinomial\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/3f1RFTIw2Ng?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Factoring a Trinomial with Leading Coefficient 1<\/h2>\n<p>Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].<\/p>\n<p>Trinomials of the form [latex]{x}^{2}+bx+c[\/latex] can be factored by finding two numbers with a product of [latex]c[\/latex] and a sum of [latex]b[\/latex]. The trinomial [latex]{x}^{2}+10x+16[\/latex], for example, can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex] because the product of those numbers is [latex]16[\/latex] and their sum is [latex]10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Factoring a Trinomial with Leading Coefficient 1<\/h3>\n<p>A trinomial of the form [latex]{x}^{2}+bx+c[\/latex] can be written in factored form as [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex] where [latex]pq=c[\/latex] and [latex]p+q=b[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Can every trinomial be factored as a product of binomials?<\/h3>\n<p><em>No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it.<\/h3>\n<ol>\n<li>List factors of [latex]c[\/latex].<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Trinomial with Leading Coefficient 1<\/h3>\n<p>Factor [latex]{x}^{2}+2x - 15[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q88306\">Solution<\/span><\/p>\n<div id=\"q88306\" class=\"hidden-answer\" style=\"display: none\">\nWe have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\n<thead>\n<tr>\n<th>Factors of [latex]-15[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-15[\/latex]<\/td>\n<td>[latex]-14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,15[\/latex]<\/td>\n<td>14<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-5[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,5[\/latex]<\/td>\n<td>2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Does the order of the factors matter?<\/h3>\n<p><em>No. Multiplication is commutative, so the order of the factors does not matter.<\/em><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor [latex]{x}^{2}-7x+6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q823058\">Solution<\/span><\/p>\n<div id=\"q823058\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(x - 6\\right)\\left(x - 1\\right)[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7897&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Factoring by Grouping<\/h2>\n<p>Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can <strong>factor by grouping<\/strong> by dividing the <em>x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Factor by Grouping<\/h3>\n<p>To factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [latex]b[\/latex]. We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression.<\/p>\n<\/div>\n<div class=\"textbox\">\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 exercises\">\n<h3>Example: Factoring a Trinomial by Grouping<\/h3>\n<p>Factor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q806328\">Solution<\/span><\/p>\n<div id=\"q806328\" class=\"hidden-answer\" style=\"display: none\">\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.<\/p>\n<table summary=\"A table with 7 rows and 2 columns. The first column is labeled: Factors of -30 while the second is labeled: Sum of Factors. The entries in the first column are: 1, -30; -1, 30; 2, -15; -2, 15; 3, -10; and -3, 10. The entries in the second column are: -29, 29, -13, 13, -7, and 7.\">\n<thead>\n<tr>\n<th>Factors of [latex]-30[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,-30[\/latex]<\/td>\n<td>[latex]-29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1,30[\/latex]<\/td>\n<td>29<\/td>\n<\/tr>\n<tr>\n<td>[latex]2,-15[\/latex]<\/td>\n<td>[latex]-13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,15[\/latex]<\/td>\n<td>13<\/td>\n<\/tr>\n<tr>\n<td>[latex]3,-10[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3,10[\/latex]<\/td>\n<td>7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>So [latex]p=-3[\/latex] and [latex]q=10[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill & \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill & \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill & \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<h4>Analysis of the Solution<\/h4>\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><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor the following.<\/p>\n<ol>\n<li>[latex]2{x}^{2}+9x+9[\/latex]<\/li>\n<li>[latex]6{x}^{2}+x - 1[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q343485\">Solution<\/span><\/p>\n<div id=\"q343485\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\left(2x+3\\right)\\left(x+3\\right)[\/latex]<\/li>\n<li>[latex]\\left(3x - 1\\right)\\left(2x+1\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7908&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>In the next video we show another example of how to factor a trinomial by grouping.<br \/>\n<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\/agDaQ_cZnNc?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-1237\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Example: Greatest Common Factor. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/3f1RFTIw2Ng\">https:\/\/youtu.be\/3f1RFTIw2Ng<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Factoring Trinomials by Grouping. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <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>Question ID 7886, 7897, 7908. <strong>Authored by<\/strong>: Tyler Wallace. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: OpenStax College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/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":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"OpenStax College Algebra\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and 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