{"id":59,"date":"2019-02-06T17:31:09","date_gmt":"2019-02-06T17:31:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/coursedemo-collegealgebracoreq\/chapter\/factoring-basics\/"},"modified":"2019-02-12T21:42:13","modified_gmt":"2019-02-12T21:42:13","slug":"factoring-basics","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/chapter\/factoring-basics\/","title":{"raw":"Factoring Basics","rendered":"Factoring Basics"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify and factor the greatest common factor 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\nRecall that the <strong>greatest common factor<\/strong> (GCF) of two numbers is the largest number that divides evenly into both numbers. For example, [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\nFinding and factoring out a GCF from a polynomial is the first skill involved in factoring polynomials.\r\n<h2>Factoring a GCF out of a polynomial<\/h2>\r\nWhen factoring a polynomial expression, our first step is to check to see if each term contains a common factor. If so, we factor out the greatest amount we can from each term. To make it less challenging to find this GCF of the polynomial terms, first 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 a polynomial is the largest polynomial that divides evenly into each term of the polynomial.\r\n\r\n<\/div>\r\nTo factor out a GCF from a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property \"backwards\" to rewrite the polynomial in a factored form. Recall that the <strong>distributive property of multiplication over addition<\/strong> states that a product of a number and a sum is the same as the sum of the products.\r\n<div class=\"textbox shaded\">\r\n<h4>Distributive Property Forward and Backward<\/h4>\r\n<p style=\"text-align: left;\">Forward:\u00a0<em>We distribute [latex]a[\/latex] over [latex]b+c[\/latex]<\/em>.<\/p>\r\n<p style=\"text-align: center;\">[latex]a\\left(b+c\\right)=ab+ac[\/latex].<\/p>\r\nBackward:\u00a0<em>We factor [latex]a[\/latex] out of [latex]ab+ac[\/latex].<\/em>\r\n<p style=\"text-align: center;\">[latex]ab+ac=a\\left(b+c\\right)[\/latex].<\/p>\r\nWe have seen that we can distribute a factor over a sum or difference. Now we see that we can \"undo\" the distributive property with factoring.\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]25b^{3}+10b^{2}[\/latex].\r\n\r\n[reveal-answer q=\"716902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"716902\"]Find the GCF.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,25b^{3}=5\\cdot5\\cdot{b}\\cdot{b}\\cdot{b}\\\\\\,\\,10b^{2}=5\\cdot2\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5\\cdot{b}\\cdot{b}=5b^{2}\\end{array}[\/latex]<\/p>\r\nRewrite each term with the GCF as one factor.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3} = 5b^{2}\\cdot5b\\\\10b^{2}=5b^{2}\\cdot2\\end{array}[\/latex]<\/p>\r\nRewrite the polynomial using the factored terms in place of the original terms.\r\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b\\right)+5b^{2}\\left(2\\right)[\/latex]<\/p>\r\nFactor out the [latex]5b^{2}[\/latex].\r\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b+2\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]5b^{2}\\left(5b+2\\right)[\/latex]\r\n<h4>Analysis<\/h4>\r\nThe factored form of the polynomial [latex]25b^{3}+10b^{2}[\/latex] is [latex]5b^{2}\\left(5b+2\\right)[\/latex]. You can check this by doing the multiplication. [latex]5b^{2}\\left(5b+2\\right)=25b^{3}+10b^{2}[\/latex].\r\n\r\nNote that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.\r\n\r\nFor example:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3}+10b^{2}=5\\left(5b^{3}+2b^{2}\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }5.\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5b^{2}\\left(5b+2\\right) \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }b^{2}.\\end{array}[\/latex]<\/p>\r\nNotice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we see two more examples of how to find and factor the GCF from binomials.\r\n\r\nhttps:\/\/youtu.be\/25_f_mVab_4\r\n<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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"113189\"]\r\n\r\nFirst find the GCF of the expression. The GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. The GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions of the form [latex]{x}^{n}[\/latex] will always be the lowest exponent.) 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\nThe GCF may not always be a monomial. Here is an example of a GCF that is a binomial.\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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"94532\"]\r\n\r\n[latex]\\left({b}^{2}-a\\right)\\left(x+6\\right)[\/latex][\/hidden-answer]\r\n[ohm_question]7888[\/ohm_question]\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 these 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<strong>Can every trinomial be factored as a product of binomials?<\/strong>\r\n\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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"88306\"]\r\n\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>[latex]14[\/latex]<\/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>[latex]2[\/latex]<\/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<strong>Does the order of the factors matter?<\/strong>\r\n\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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"823058\"]\r\n\r\n[latex]\\left(x - 6\\right)\\left(x - 1\\right)[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]7897[\/ohm_question]\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: Factoring by Grouping<\/h3>\r\nTo factor a trinomial of 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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"806328\"]\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<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 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\"]Show 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[ohm_question]7908[\/ohm_question]\r\n\r\n[ohm_question]1846[\/ohm_question]\r\n\r\n<\/div>\r\nIn the next video we see another example of how to factor a trinomial by grouping.\r\n\r\n[embed]https:\/\/youtu.be\/agDaQ_cZnNc[\/embed]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify and factor the greatest common factor 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>Recall that the <strong>greatest common factor<\/strong> (GCF) of two numbers is the largest number that divides evenly into both numbers. For example, [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>Finding and factoring out a GCF from a polynomial is the first skill involved in factoring polynomials.<\/p>\n<h2>Factoring a GCF out of a polynomial<\/h2>\n<p>When factoring a polynomial expression, our first step is to check to see if each term contains a common factor. If so, we factor out the greatest amount we can from each term. To make it less challenging to find this GCF of the polynomial terms, first 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 a polynomial is the largest polynomial that divides evenly into each term of the polynomial.<\/p>\n<\/div>\n<p>To factor out a GCF from a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property &#8220;backwards&#8221; to rewrite the polynomial in a factored form. Recall that the <strong>distributive property of multiplication over addition<\/strong> states that a product of a number and a sum is the same as the sum of the products.<\/p>\n<div class=\"textbox shaded\">\n<h4>Distributive Property Forward and Backward<\/h4>\n<p style=\"text-align: left;\">Forward:\u00a0<em>We distribute [latex]a[\/latex] over [latex]b+c[\/latex]<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]a\\left(b+c\\right)=ab+ac[\/latex].<\/p>\n<p>Backward:\u00a0<em>We factor [latex]a[\/latex] out of [latex]ab+ac[\/latex].<\/em><\/p>\n<p style=\"text-align: center;\">[latex]ab+ac=a\\left(b+c\\right)[\/latex].<\/p>\n<p>We have seen that we can distribute a factor over a sum or difference. Now we see that we can &#8220;undo&#8221; the distributive property with factoring.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]25b^{3}+10b^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q716902\">Show Solution<\/span><\/p>\n<div id=\"q716902\" class=\"hidden-answer\" style=\"display: none\">Find the GCF.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,25b^{3}=5\\cdot5\\cdot{b}\\cdot{b}\\cdot{b}\\\\\\,\\,10b^{2}=5\\cdot2\\cdot{b}\\cdot{b}\\\\\\text{GCF}=5\\cdot{b}\\cdot{b}=5b^{2}\\end{array}[\/latex]<\/p>\n<p>Rewrite each term with the GCF as one factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3} = 5b^{2}\\cdot5b\\\\10b^{2}=5b^{2}\\cdot2\\end{array}[\/latex]<\/p>\n<p>Rewrite the polynomial using the factored terms in place of the original terms.<\/p>\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b\\right)+5b^{2}\\left(2\\right)[\/latex]<\/p>\n<p>Factor out the [latex]5b^{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]5b^{2}\\left(5b+2\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]5b^{2}\\left(5b+2\\right)[\/latex]<\/p>\n<h4>Analysis<\/h4>\n<p>The factored form of the polynomial [latex]25b^{3}+10b^{2}[\/latex] is [latex]5b^{2}\\left(5b+2\\right)[\/latex]. You can check this by doing the multiplication. [latex]5b^{2}\\left(5b+2\\right)=25b^{3}+10b^{2}[\/latex].<\/p>\n<p>Note that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.<\/p>\n<p>For example:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}25b^{3}+10b^{2}=5\\left(5b^{3}+2b^{2}\\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }5.\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5b^{2}\\left(5b+2\\right) \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Factor out }b^{2}.\\end{array}[\/latex]<\/p>\n<p>Notice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we see two more examples of how to find and factor the GCF from binomials.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 1:  Identify GCF and Factor a Binomial\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/25_f_mVab_4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<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\">Show Solution<\/span><\/p>\n<div id=\"q113189\" class=\"hidden-answer\" style=\"display: none\">\n<p>First find the GCF of the expression. The GCF of [latex]6,45[\/latex], and [latex]21[\/latex] is [latex]3[\/latex]. The GCF of [latex]{x}^{3},{x}^{2}[\/latex], and [latex]x[\/latex] is [latex]x[\/latex]. (Note that the GCF of a set of expressions of the form [latex]{x}^{n}[\/latex] will always be the lowest exponent.) 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<p>The GCF may not always be a monomial. Here is an example of a GCF that is a binomial.<\/p>\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\">Show Solution<\/span><\/p>\n<div id=\"q94532\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left({b}^{2}-a\\right)\\left(x+6\\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm7888\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7888&theme=oea&iframe_resize_id=ohm7888&show_question_numbers\" width=\"100%\" height=\"150\"><\/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-3\" 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 these 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<p><strong>Can every trinomial be factored as a product of binomials?<\/strong><\/p>\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\">Show Solution<\/span><\/p>\n<div id=\"q88306\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex]. In the table, we list factors until we find a pair with the desired sum.<\/p>\n<table 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>[latex]14[\/latex]<\/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>[latex]2[\/latex]<\/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<p><strong>Does the order of the factors matter?<\/strong><\/p>\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\">Show Solution<\/span><\/p>\n<div id=\"q823058\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(x - 6\\right)\\left(x - 1\\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm7897\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7897&theme=oea&iframe_resize_id=ohm7897&show_question_numbers\" width=\"100%\" height=\"150\"><\/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: Factoring by Grouping<\/h3>\n<p>To factor a trinomial of 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\">Show Solution<\/span><\/p>\n<div id=\"q806328\" 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<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 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\">Show 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=\"ohm7908\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7908&theme=oea&iframe_resize_id=ohm7908&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm1846\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1846&theme=oea&iframe_resize_id=ohm1846&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the next video we see another example of how to factor a trinomial by grouping.<\/p>\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\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-59\">\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":17533,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"OpenStax College 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