{"id":1755,"date":"2023-10-12T00:32:06","date_gmt":"2023-10-12T00:32:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-factoring-polynomials\/"},"modified":"2023-10-12T00:32:06","modified_gmt":"2023-10-12T00:32:06","slug":"introduction-factoring-polynomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-factoring-polynomials\/","title":{"raw":"Factoring Polynomials","rendered":"Factoring Polynomials"},"content":{"raw":"\n\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Identify and factor the greatest common factor of a polynomial.<\/li>\n \t<li>Factor a trinomial with leading coefficient 1.<\/li>\n \t<li>Factor by grouping.<\/li>\n \t<li>Factor a perfect square trinomial.<\/li>\n \t<li>Factor a difference of squares.<\/li>\n \t<li>Factor a sum and difference of cubes.<\/li>\n \t<li>Factor an expression with negative or fractional exponents.<\/li>\n<\/ul>\n<\/div>\nImagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in the figure below.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21210700\/CNX_CAT_Figure_01_05_001.jpg\" alt=\"A large rectangle with smaller squares and a rectangle inside. The length of the outer rectangle is 6x and the width is 10x. The side length of the squares is 4 and the height of the width of the inner rectangle is 4.\" width=\"487\" height=\"259\">\n\nThe area of the entire region can be found using the formula for the area of a rectangle.\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill A&amp; =&amp; lw\\hfill \\\\ &amp; =&amp; 10x\\cdot 6x\\hfill \\\\ &amp; =&amp; 60{x}^{2}{\\text{ units}}^{2}\\hfill \\end{array}[\/latex]<\/p>\nThe areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of [latex]A={s}^{2}={4}^{2}=16[\/latex] units<sup>2<\/sup>. The other rectangular region has one side of length [latex]10x - 8[\/latex] and one side of length [latex]4[\/latex], giving an area of [latex]A=lw=4\\left(10x - 8\\right)=40x - 32[\/latex] units<sup>2<\/sup>. So the region that must be subtracted has an area of [latex]2\\left(16\\right)+40x - 32=40x[\/latex] units<sup>2<\/sup>.\n\nThe area of the region that requires grass seed is found by subtracting [latex]60{x}^{2}-40x[\/latex] units<sup>2<\/sup>. This area can also be expressed in factored form as [latex]20x\\left(3x - 2\\right)[\/latex] units<sup>2<\/sup>. We can confirm that this is an equivalent expression by multiplying.\n\nMany polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions.\n<h2>Factoring Basics<\/h2>\nWhen we studied fractions, we learned 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].\n\nWhen factoring a polynomial expression, our first step is to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.\n<div class=\"textbox\">\n<h3>A General Note: Greatest Common Factor<\/h3>\nThe <strong>greatest common factor<\/strong> (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials.\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a polynomial expression, factor out the greatest common factor<strong>\n<\/strong><\/h3>\n<ol>\n \t<li>Identify the GCF of the coefficients.<\/li>\n \t<li>Identify the GCF of the variables.<\/li>\n \t<li>Combine to find the GCF of the expression.<\/li>\n \t<li>Determine what the GCF needs to be multiplied by to obtain each term in the expression.<\/li>\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>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring the Greatest Common Factor<\/h3>\nFactor [latex]6{x}^{3}{y}^{3}+45{x}^{2}{y}^{2}+21xy[\/latex].\n\n[reveal-answer q=\"113189\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"113189\"]\n\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].\n\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].\n\nFinally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by.\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>\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].\n\n<\/div>\n<div>[\/hidden-answer]<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFactor [latex]x\\left({b}^{2}-a\\right)+6\\left({b}^{2}-a\\right)[\/latex] by pulling out the GCF.\n\n[reveal-answer q=\"94532\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"94532\"]\n\n[latex]\\left({b}^{2}-a\\right)\\left(x+6\\right)[\/latex][\/hidden-answer]\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7886&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\n\n<\/div>\n<h3>Factoring a Trinomial with Leading Coefficient 1<\/h3>\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].\n\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].\n<div class=\"textbox\">\n<h3>A General Note: Factoring a Trinomial with Leading Coefficient 1<\/h3>\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].\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<strong>Can every trinomial be factored as a product of binomials?<\/strong>\n\n<em>No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/em>\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a trinomial OF the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\n<ol>\n \t<li>List factors of [latex]c[\/latex].<\/li>\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>\n \t<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>\nFactor [latex]{x}^{2}+2x - 15[\/latex].\n\n[reveal-answer q=\"88306\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"88306\"]\n\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.\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>\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].\n<h4>Analysis of the Solution<\/h4>\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].\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<strong>Does the order of the factors matter?<\/strong>\n\n<em>No. Multiplication is commutative, so the order of the factors does not matter.<\/em>\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFactor [latex]{x}^{2}-7x+6[\/latex].\n\n[reveal-answer q=\"823058\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"823058\"]\n\n[latex]\\left(x - 6\\right)\\left(x - 1\\right)[\/latex][\/hidden-answer]\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7897&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\n\n<\/div>\n<h3>Factoring by Grouping<\/h3>\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.\n<div class=\"textbox\">\n<h3>A General Note: Factoring by Grouping<\/h3>\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.\n\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 \t<li>List factors of [latex]ac[\/latex].<\/li>\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>\n \t<li>Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\n \t<li>Pull out the GCF of [latex]a{x}^{2}+px[\/latex].<\/li>\n \t<li>Pull out the GCF of [latex]qx+c[\/latex].<\/li>\n \t<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>\nFactor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.\n\n[reveal-answer q=\"806328\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"806328\"]\n\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.\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>\nSo [latex]p=-3[\/latex] and [latex]q=10[\/latex].\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>\n<div>\n<h4>Analysis of the Solution<\/h4>\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].\n\n<\/div>\n<div>[\/hidden-answer]<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFactor the following.\n<ol>\n \t<li>[latex]2{x}^{2}+9x+9[\/latex]<\/li>\n \t<li>[latex]6{x}^{2}+x - 1[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"343485\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"343485\"]\n<ol>\n \t<li>[latex]\\left(2x+3\\right)\\left(x+3\\right)[\/latex]<\/li>\n \t<li>[latex]\\left(3x - 1\\right)\\left(2x+1\\right)[\/latex]<\/li>\n<\/ol>\n[\/hidden-answer]\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7908&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\n\n<\/div>\n<h2>Factoring Special Cases<\/h2>\n<h3>Factoring a Perfect Square Trinomial<\/h3>\nA perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}&amp; =&amp; {\\left(a+b\\right)}^{2}\\hfill \\\\ &amp; \\text{and}&amp; \\\\ \\hfill {a}^{2}-2ab+{b}^{2}&amp; =&amp; {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div style=\"text-align: left;\">We can use this equation to factor any perfect square trinomial.<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Perfect Square Trinomials<\/h3>\nA perfect square trinomial can be written as the square of a binomial:\n<div style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a perfect square trinomial, factor it into the square of a binomial<strong>\n<\/strong><\/h3>\n<ol>\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\n \t<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\n \t<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Perfect Square Trinomial<\/h3>\nFactor [latex]25{x}^{2}+20x+4[\/latex].\n\n[reveal-answer q=\"114092\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"114092\"]\n\nNotice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex]. Then check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(5x+2\\right)}^{2}[\/latex].\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFactor [latex]49{x}^{2}-14x+1[\/latex].\n\n[reveal-answer q=\"530221\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"530221\"]\n\n[latex]{\\left(7x - 1\\right)}^{2}[\/latex][\/hidden-answer]\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7919&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\n\n<\/div>\n<h3>Factoring a Difference of Squares<\/h3>\nA difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n[latex]\\\\[\/latex]\n\nWe can use this equation to factor any differences of squares.\n<div class=\"textbox\">\n<h3>A General Note: Differences of Squares<\/h3>\nA difference of squares can be rewritten as two factors containing the same terms but opposite signs.\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a difference of squares, factor it into binomials<\/h3>\n<ol>\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\n \t<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Difference of Squares<\/h3>\nFactor [latex]9{x}^{2}-25[\/latex].\n\n[reveal-answer q=\"830417\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"830417\"]\n\nNotice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex]. The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFactor [latex]81{y}^{2}-100[\/latex].\n\n[reveal-answer q=\"979538\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"979538\"]\n\n[latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex]\n\n[\/hidden-answer]\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7929&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<strong>Is there a formula to factor the sum of squares?<\/strong>\n\n<em>No. A sum of squares cannot be factored.<\/em>\n\n<\/div>\n<h3>Factoring the Sum and Difference of Cubes<\/h3>\nNow we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\n[latex]\\\\[\/latex]\n\nSimilarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\n[latex]\\\\[\/latex]\n\nWe can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: <strong>S<\/strong>ame <strong>O<\/strong>pposite <strong>A<\/strong>lways <strong>P<\/strong>ositive. For example, consider the following example.\n<div style=\"text-align: center;\">[latex]{x}^{3}-{2}^{3}=\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)[\/latex]<\/div>\nThe sign of the first 2 is the <em>same<\/em> as the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. The sign of the [latex]2x[\/latex] term is <em>opposite<\/em> the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. And the sign of the last term, 4, is <em>always positive<\/em>.\n<div class=\"textbox\">\n<h3>A General Note: Sum and Difference of Cubes<\/h3>\nWe can factor the sum of two cubes as\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\nWe can factor the difference of two cubes as\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a sum of cubes or difference of cubes, factor it<\/h3>\n<ol>\n \t<li>Confirm that the first and last term are cubes, [latex]{a}^{3}+{b}^{3}[\/latex] or [latex]{a}^{3}-{b}^{3}[\/latex].<\/li>\n \t<li>For a sum of cubes, write the factored form as [latex]\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]. For a difference of cubes, write the factored form as [latex]\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Sum of Cubes<\/h3>\nFactor [latex]{x}^{3}+512[\/latex].\n\n[reveal-answer q=\"22673\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"22673\"]\n\nNotice that [latex]{x}^{3}[\/latex] and [latex]512[\/latex] are cubes because [latex]{8}^{3}=512[\/latex]. Rewrite the sum of cubes as [latex]\\left(x+8\\right)\\left({x}^{2}-8x+64\\right)[\/latex].\n<h4>Analysis of the Solution<\/h4>\nAfter writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFactor the sum of cubes [latex]216{a}^{3}+{b}^{3}[\/latex].\n\n[reveal-answer q=\"496204\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"496204\"]\n\n[latex]\\left(6a+b\\right)\\left(36{a}^{2}-6ab+{b}^{2}\\right)[\/latex][\/hidden-answer]\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Difference of Cubes<\/h3>\nFactor [latex]8{x}^{3}-125[\/latex].\n\n[reveal-answer q=\"741836\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"741836\"]\n\nNotice that [latex]8{x}^{3}[\/latex] and [latex]125[\/latex] are cubes because [latex]8{x}^{3}={\\left(2x\\right)}^{3}[\/latex] and [latex]125={5}^{3}[\/latex]. Write the difference of cubes as [latex]\\left(2x - 5\\right)\\left(4{x}^{2}+10x+25\\right)[\/latex].\n<h4>Analysis of the Solution<\/h4>\nJust as with the sum of cubes, we will not be able to further factor the trinomial portion.\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFactor the difference of cubes: [latex]1,000{x}^{3}-1[\/latex].\n\n[reveal-answer q=\"510077\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"510077\"]\n\n[latex]\\left(10x - 1\\right)\\left(100{x}^{2}+10x+1\\right)[\/latex][\/hidden-answer]\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7922&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\n\n<\/div>\n<h3>Factoring Expressions with Fractional or Negative Exponents<\/h3>\nExpressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, [latex]2{x}^{\\frac{1}{4}}+5{x}^{\\frac{3}{4}}[\/latex] can be factored by pulling out [latex]{x}^{\\frac{1}{4}}[\/latex] and being rewritten as [latex]{x}^{\\frac{1}{4}}\\left(2+5{x}^{\\frac{1}{2}}\\right)[\/latex].\n<div class=\"textbox exercises\">\n<h3>Example: Factoring an Expression with Fractional or Negative Exponents<\/h3>\nFactor [latex]3x{\\left(x+2\\right)}^{\\frac{-1}{3}}+4{\\left(x+2\\right)}^{\\frac{2}{3}}[\/latex].\n\n[reveal-answer q=\"164967\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"164967\"]\n\nFactor out the term with the lowest value of the exponent. In this case, that would be [latex]{\\left(x+2\\right)}^{-\\frac{1}{3}}[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4\\left(x+2\\right)\\right)\\hfill &amp; \\text{Factor out the GCF}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4x+8\\right)\\hfill &amp; \\text{Simplify}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(7x+8\\right)\\hfill &amp; \\end{array}[\/latex]<\/div>\n<div>[\/hidden-answer]<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFactor [latex]2{\\left(5a - 1\\right)}^{\\frac{3}{4}}+7a{\\left(5a - 1\\right)}^{-\\frac{1}{4}}[\/latex].\n\n[reveal-answer q=\"989124\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"989124\"]\n\n[latex]{\\left(5a - 1\\right)}^{-\\frac{1}{4}}\\left(17a - 2\\right)[\/latex][\/hidden-answer]\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93666&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93668&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe>\n\n<\/div>\n<h2>Key Equations<\/h2>\n<table>\n<tbody>\n<tr>\n<td><strong>difference of squares<\/strong><\/td>\n<td>[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>perfect square trinomial<\/strong><\/td>\n<td>[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>sum of cubes<\/strong><\/td>\n<td>[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>difference of cubes<\/strong><\/td>\n<td>[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n \t<li>The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem.<\/li>\n \t<li>Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term.<\/li>\n \t<li>Trinomials can be factored using a process called factoring by grouping.<\/li>\n \t<li>Perfect square trinomials and the difference of squares are special products and can be factored using equations.<\/li>\n \t<li>The sum of cubes and the difference of cubes can be factored using equations.<\/li>\n \t<li>Polynomials containing fractional and negative exponents can be factored by pulling out a GCF.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165133085661\" class=\"definition\">\n \t<dt>\n<dl id=\"fs-id1165133085661\" class=\"definition\">\n \t<dt>\n<dl id=\"fs-id1165133085661\" class=\"definition\">\n \t<dt><strong>factor by grouping<\/strong><\/dt>\n \t<dd id=\"fs-id1165133085665\">a method for factoring a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] 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<\/dd>\n<\/dl>\n<\/dt>\n<\/dl>\n<\/dt>\n \t<dt>\n<dl id=\"fs-id1165133085661\" class=\"definition\">\n \t<dt>\n<dl id=\"fs-id1165137644987\" class=\"definition\">\n \t<dt><strong>greatest common factor<\/strong><\/dt>\n \t<dd id=\"fs-id1165137644990\">the largest polynomial that divides evenly into each polynomial<\/dd>\n<\/dl>\n<\/dt>\n<\/dl>\n<\/dt>\n<\/dl>\n\n","rendered":"<div class=\"bcc-box bcc-highlight\">\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<li>Factor a perfect square trinomial.<\/li>\n<li>Factor a difference of squares.<\/li>\n<li>Factor a sum and difference of cubes.<\/li>\n<li>Factor an expression with negative or fractional exponents.<\/li>\n<\/ul>\n<\/div>\n<p>Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in the figure below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21210700\/CNX_CAT_Figure_01_05_001.jpg\" alt=\"A large rectangle with smaller squares and a rectangle inside. The length of the outer rectangle is 6x and the width is 10x. The side length of the squares is 4 and the height of the width of the inner rectangle is 4.\" width=\"487\" height=\"259\" \/><\/p>\n<p>The area of the entire region can be found using the formula for the area of a rectangle.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill A& =& lw\\hfill \\\\ & =& 10x\\cdot 6x\\hfill \\\\ & =& 60{x}^{2}{\\text{ units}}^{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of [latex]A={s}^{2}={4}^{2}=16[\/latex] units<sup>2<\/sup>. The other rectangular region has one side of length [latex]10x - 8[\/latex] and one side of length [latex]4[\/latex], giving an area of [latex]A=lw=4\\left(10x - 8\\right)=40x - 32[\/latex] units<sup>2<\/sup>. So the region that must be subtracted has an area of [latex]2\\left(16\\right)+40x - 32=40x[\/latex] units<sup>2<\/sup>.<\/p>\n<p>The area of the region that requires grass seed is found by subtracting [latex]60{x}^{2}-40x[\/latex] units<sup>2<\/sup>. This area can also be expressed in factored form as [latex]20x\\left(3x - 2\\right)[\/latex] units<sup>2<\/sup>. We can confirm that this is an equivalent expression by multiplying.<\/p>\n<p>Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions.<\/p>\n<h2>Factoring Basics<\/h2>\n<p>When we studied fractions, we learned 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 is 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\">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<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=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7886&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Factoring a Trinomial with Leading Coefficient 1<\/h3>\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 OF 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 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=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7897&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Factoring by Grouping<\/h3>\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=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7908&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Factoring Special Cases<\/h2>\n<h3>Factoring a Perfect Square Trinomial<\/h3>\n<p>A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}& =& {\\left(a+b\\right)}^{2}\\hfill \\\\ & \\text{and}& \\\\ \\hfill {a}^{2}-2ab+{b}^{2}& =& {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div style=\"text-align: left;\">We can use this equation to factor any perfect square trinomial.<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Perfect Square Trinomials<\/h3>\n<p>A perfect square trinomial can be written as the square of a binomial:<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a perfect square trinomial, factor it into the square of a binomial<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\n<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Perfect Square Trinomial<\/h3>\n<p>Factor [latex]25{x}^{2}+20x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q114092\">Show Solution<\/span><\/p>\n<div id=\"q114092\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex]. Then check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(5x+2\\right)}^{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor [latex]49{x}^{2}-14x+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q530221\">Show Solution<\/span><\/p>\n<div id=\"q530221\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\left(7x - 1\\right)}^{2}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7919&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Factoring a Difference of Squares<\/h3>\n<p>A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<p>[latex]\\\\[\/latex]<\/p>\n<p>We can use this equation to factor any differences of squares.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Differences of Squares<\/h3>\n<p>A difference of squares can be rewritten as two factors containing the same terms but opposite signs.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a difference of squares, factor it into binomials<\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Difference of Squares<\/h3>\n<p>Factor [latex]9{x}^{2}-25[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q830417\">Show Solution<\/span><\/p>\n<div id=\"q830417\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex]. The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor [latex]81{y}^{2}-100[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q979538\">Show Solution<\/span><\/p>\n<div id=\"q979538\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7929&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Is there a formula to factor the sum of squares?<\/strong><\/p>\n<p><em>No. A sum of squares cannot be factored.<\/em><\/p>\n<\/div>\n<h3>Factoring the Sum and Difference of Cubes<\/h3>\n<p>Now we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\n<p>[latex]\\\\[\/latex]<\/p>\n<p>Similarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\n<p>[latex]\\\\[\/latex]<\/p>\n<p>We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: <strong>S<\/strong>ame <strong>O<\/strong>pposite <strong>A<\/strong>lways <strong>P<\/strong>ositive. For example, consider the following example.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{3}-{2}^{3}=\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)[\/latex]<\/div>\n<p>The sign of the first 2 is the <em>same<\/em> as the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. The sign of the [latex]2x[\/latex] term is <em>opposite<\/em> the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. And the sign of the last term, 4, is <em>always positive<\/em>.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Sum and Difference of Cubes<\/h3>\n<p>We can factor the sum of two cubes as<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\n<p>We can factor the difference of two cubes as<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a sum of cubes or difference of cubes, factor it<\/h3>\n<ol>\n<li>Confirm that the first and last term are cubes, [latex]{a}^{3}+{b}^{3}[\/latex] or [latex]{a}^{3}-{b}^{3}[\/latex].<\/li>\n<li>For a sum of cubes, write the factored form as [latex]\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]. For a difference of cubes, write the factored form as [latex]\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Sum of Cubes<\/h3>\n<p>Factor [latex]{x}^{3}+512[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q22673\">Show Solution<\/span><\/p>\n<div id=\"q22673\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]{x}^{3}[\/latex] and [latex]512[\/latex] are cubes because [latex]{8}^{3}=512[\/latex]. Rewrite the sum of cubes as [latex]\\left(x+8\\right)\\left({x}^{2}-8x+64\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor the sum of cubes [latex]216{a}^{3}+{b}^{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q496204\">Show Solution<\/span><\/p>\n<div id=\"q496204\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(6a+b\\right)\\left(36{a}^{2}-6ab+{b}^{2}\\right)[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring a Difference of Cubes<\/h3>\n<p>Factor [latex]8{x}^{3}-125[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q741836\">Show Solution<\/span><\/p>\n<div id=\"q741836\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]8{x}^{3}[\/latex] and [latex]125[\/latex] are cubes because [latex]8{x}^{3}={\\left(2x\\right)}^{3}[\/latex] and [latex]125={5}^{3}[\/latex]. Write the difference of cubes as [latex]\\left(2x - 5\\right)\\left(4{x}^{2}+10x+25\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Just as with the sum of cubes, we will not be able to further factor the trinomial portion.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor the difference of cubes: [latex]1,000{x}^{3}-1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q510077\">Show Solution<\/span><\/p>\n<div id=\"q510077\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(10x - 1\\right)\\left(100{x}^{2}+10x+1\\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7922&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Factoring Expressions with Fractional or Negative Exponents<\/h3>\n<p>Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, [latex]2{x}^{\\frac{1}{4}}+5{x}^{\\frac{3}{4}}[\/latex] can be factored by pulling out [latex]{x}^{\\frac{1}{4}}[\/latex] and being rewritten as [latex]{x}^{\\frac{1}{4}}\\left(2+5{x}^{\\frac{1}{2}}\\right)[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring an Expression with Fractional or Negative Exponents<\/h3>\n<p>Factor [latex]3x{\\left(x+2\\right)}^{\\frac{-1}{3}}+4{\\left(x+2\\right)}^{\\frac{2}{3}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q164967\">Show Solution<\/span><\/p>\n<div id=\"q164967\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factor out the term with the lowest value of the exponent. In this case, that would be [latex]{\\left(x+2\\right)}^{-\\frac{1}{3}}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4\\left(x+2\\right)\\right)\\hfill & \\text{Factor out the GCF}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4x+8\\right)\\hfill & \\text{Simplify}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(7x+8\\right)\\hfill & \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor [latex]2{\\left(5a - 1\\right)}^{\\frac{3}{4}}+7a{\\left(5a - 1\\right)}^{-\\frac{1}{4}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q989124\">Show Solution<\/span><\/p>\n<div id=\"q989124\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\left(5a - 1\\right)}^{-\\frac{1}{4}}\\left(17a - 2\\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93666&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=93668&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Key Equations<\/h2>\n<table>\n<tbody>\n<tr>\n<td><strong>difference of squares<\/strong><\/td>\n<td>[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>perfect square trinomial<\/strong><\/td>\n<td>[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>sum of cubes<\/strong><\/td>\n<td>[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>difference of cubes<\/strong><\/td>\n<td>[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem.<\/li>\n<li>Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term.<\/li>\n<li>Trinomials can be factored using a process called factoring by grouping.<\/li>\n<li>Perfect square trinomials and the difference of squares are special products and can be factored using equations.<\/li>\n<li>The sum of cubes and the difference of cubes can be factored using equations.<\/li>\n<li>Polynomials containing fractional and negative exponents can be factored by pulling out a GCF.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165133085661\" class=\"definition\">\n<dt>\n<\/dt>\n<dt>\n<\/dt>\n<dt><strong>factor by grouping<\/strong><\/dt>\n<dd id=\"fs-id1165133085665\">a method for factoring a trinomial of the form [latex]a{x}^{2}+bx+c[\/latex] 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<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133085661\" class=\"definition\">\n<dt>\n<\/dt>\n<dt><strong>greatest common factor<\/strong><\/dt>\n<dd id=\"fs-id1165137644990\">the largest polynomial that divides evenly into each polynomial<\/dd>\n<\/dl>\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-1755\">\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, 7919, 7929, 7922. <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><li>Examples: Factoring Binomials (Special). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Li9IBp5HrFA\">https:\/\/youtu.be\/Li9IBp5HrFA<\/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>Ex1: Factor a Sum or Difference of Cubes. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tFSEpOB262M%20\">https:\/\/youtu.be\/tFSEpOB262M%20<\/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>Ex3: Factor a Sum or Difference of Cubes. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/J_0ctMrl5_0\">https:\/\/youtu.be\/J_0ctMrl5_0<\/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 93666, 93668. <strong>Authored by<\/strong>: Michael Jenck. <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":708740,"menu_order":15,"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 Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Example: Greatest 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