{"id":67,"date":"2023-06-21T13:22:29","date_gmt":"2023-06-21T13:22:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/factor-special-cases\/"},"modified":"2023-07-04T04:57:34","modified_gmt":"2023-07-04T04:57:34","slug":"factor-special-cases","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/factor-special-cases\/","title":{"raw":"\u25aa   Factoring Special Cases","rendered":"\u25aa   Factoring Special Cases"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor a perfect square trinomial.<\/li>\r\n \t<li>Factor a difference of squares.<\/li>\r\n \t<li>Factor a sum and difference of cubes.<\/li>\r\n \t<li>Factor an expression with negative or fractional exponents.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Factoring a Perfect Square Trinomial<\/h2>\r\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.\r\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>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div style=\"text-align: left;\">We can use this equation to factor any perfect square trinomial.<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Perfect Square Trinomials<\/h3>\r\nA perfect square trinomial can be written as the square of a binomial:\r\n<div style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a perfect square trinomial, factor it into the square of a binomial<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\r\n \t<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Perfect Square Trinomial<\/h3>\r\nFactor [latex]25{x}^{2}+20x+4[\/latex].\r\n\r\n[reveal-answer q=\"114092\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"114092\"]\r\n\r\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].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]49{x}^{2}-14x+1[\/latex].\r\n\r\n[reveal-answer q=\"530221\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530221\"]\r\n\r\n[latex]{\\left(7x - 1\\right)}^{2}[\/latex][\/hidden-answer]\r\n\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7919&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Factoring a Difference of Squares<\/h2>\r\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.\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n[latex]\\\\[\/latex]\r\n\r\nWe can use this equation to factor any differences of squares.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Differences of Squares<\/h3>\r\nA difference of squares can be rewritten as two factors containing the same terms but opposite signs.\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a difference of squares, factor it into binomials<\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Difference of Squares<\/h3>\r\nFactor [latex]9{x}^{2}-25[\/latex].\r\n\r\n[reveal-answer q=\"830417\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"830417\"]\r\n\r\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].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]81{y}^{2}-100[\/latex].\r\n\r\n[reveal-answer q=\"979538\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979538\"]\r\n\r\n[latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7929&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Is there a formula to factor the sum of squares?<\/strong>\r\n\r\n<em>No. A sum of squares cannot be factored.<\/em>\r\n\r\n<\/div>\r\nWatch this video to see another example of how to factor a difference of squares.\r\n\r\nhttps:\/\/youtu.be\/Li9IBp5HrFA\r\n<h2>Factoring the Sum and Difference of Cubes<\/h2>\r\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.\r\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\r\n[latex]\\\\[\/latex]\r\n\r\nSimilarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.\r\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\r\n[latex]\\\\[\/latex]\r\n\r\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.\r\n<div style=\"text-align: center;\">[latex]{x}^{3}-{2}^{3}=\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)[\/latex]<\/div>\r\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>.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Sum and Difference of Cubes<\/h3>\r\nWe can factor the sum of two cubes as\r\n<div style=\"text-align: center;\">[latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\r\nWe can factor the difference of two cubes as\r\n<div style=\"text-align: center;\">[latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a sum of cubes or difference of cubes, factor it<\/h3>\r\n<ol>\r\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>\r\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>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Sum of Cubes<\/h3>\r\nFactor [latex]{x}^{3}+512[\/latex].\r\n\r\n[reveal-answer q=\"22673\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"22673\"]\r\n\r\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].\r\n<h4>Analysis of the Solution<\/h4>\r\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.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor the sum of cubes [latex]216{a}^{3}+{b}^{3}[\/latex].\r\n\r\n[reveal-answer q=\"496204\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"496204\"]\r\n\r\n[latex]\\left(6a+b\\right)\\left(36{a}^{2}-6ab+{b}^{2}\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring a Difference of Cubes<\/h3>\r\nFactor [latex]8{x}^{3}-125[\/latex].\r\n\r\n[reveal-answer q=\"741836\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"741836\"]\r\n\r\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].\r\n<h4>Analysis of the Solution<\/h4>\r\nJust as with the sum of cubes, we will not be able to further factor the trinomial portion.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor the difference of cubes: [latex]1,000{x}^{3}-1[\/latex].\r\n\r\n[reveal-answer q=\"510077\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"510077\"]\r\n\r\n[latex]\\left(10x - 1\\right)\\left(100{x}^{2}+10x+1\\right)[\/latex][\/hidden-answer]\r\n\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7922&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\nIn the following two video examples we show more binomials that can be factored as a sum or difference of cubes.\r\n\r\nhttps:\/\/youtu.be\/tFSEpOB262M\r\n\r\nhttps:\/\/youtu.be\/J_0ctMrl5_0\r\n<h2>Factoring Expressions with Fractional or Negative Exponents<\/h2>\r\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].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring an Expression with Fractional or Negative Exponents<\/h3>\r\nFactor [latex]3x{\\left(x+2\\right)}^{\\frac{-1}{3}}+4{\\left(x+2\\right)}^{\\frac{2}{3}}[\/latex].\r\n\r\n[reveal-answer q=\"164967\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164967\"]\r\n\r\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].\r\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>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div>\r\n\r\nThe following video provides more examples of how to factor expressions with negative exponents.\r\n\r\n[embed]https:\/\/youtu.be\/4w99g0GZOCk[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]2{\\left(5a - 1\\right)}^{\\frac{3}{4}}+7a{\\left(5a - 1\\right)}^{-\\frac{1}{4}}[\/latex].\r\n\r\n[reveal-answer q=\"989124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"989124\"]\r\n\r\n[latex]{\\left(5a - 1\\right)}^{-\\frac{1}{4}}\\left(17a - 2\\right)[\/latex][\/hidden-answer]\r\n\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=93666&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=93668&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\nThe following video provides more examples of how to factor expressions with fractional exponents.\r\n\r\nhttps:\/\/youtu.be\/R6BzjR2O4z8","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\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<h2>Factoring a Perfect Square Trinomial<\/h2>\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:\/\/www.myopenmath.com\/multiembedq.php?id=7919&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Factoring a Difference of Squares<\/h2>\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:\/\/www.myopenmath.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<p>Watch this video to see another example of how to factor a difference of squares.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Factor a Difference of Squares\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Li9IBp5HrFA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Factoring the Sum and Difference of Cubes<\/h2>\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:\/\/www.myopenmath.com\/multiembedq.php?id=7922&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>In the following two video examples we show more binomials that can be factored as a sum or difference of cubes.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 1:  Factor a Sum or Difference of Cubes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tFSEpOB262M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 3:  Factor a Sum or Difference of Cubes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/J_0ctMrl5_0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Factoring Expressions with Fractional or Negative Exponents<\/h2>\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>\n<p>The following video provides more examples of how to factor expressions with negative exponents.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factor Expressions with Negative Exponents\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/4w99g0GZOCk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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:\/\/www.myopenmath.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:\/\/www.myopenmath.com\/multiembedq.php?id=93668&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>The following video provides more examples of how to factor expressions with fractional exponents.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Factor Expressions with Fractional Exponents\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/R6BzjR2O4z8?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-67\">\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>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 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>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":395986,"menu_order":13,"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\":\"Examples: Factoring Binomials (Special)\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Li9IBp5HrFA\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex1: Factor a Sum or Difference of Cubes\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/tFSEpOB262M \",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex3: Factor a Sum or Difference of Cubes\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/J_0ctMrl5_0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 7919, 7929, 7922\",\"author\":\"Tyler Wallace\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License, CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 93666, 93668\",\"author\":\"Michael Jenck\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"a0483076-695b-4d9b-a196-d797c64b172c","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-67","chapter","type-chapter","status-publish","hentry"],"part":54,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/67","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/67\/revisions"}],"predecessor-version":[{"id":847,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/67\/revisions\/847"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/54"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/67\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=67"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=67"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=67"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=67"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}