{"id":63,"date":"2023-06-21T13:22:29","date_gmt":"2023-06-21T13:22:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/special-cases-of-polynomials\/"},"modified":"2023-07-04T04:56:49","modified_gmt":"2023-07-04T04:56:49","slug":"special-cases-of-polynomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/special-cases-of-polynomials\/","title":{"raw":"\u25aa   Special Cases of Polynomials","rendered":"\u25aa   Special Cases of Polynomials"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Square a binomial.<\/li>\r\n \t<li>Find a difference of squares.<\/li>\r\n \t<li>Perform operations on polynomials with several variables.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Perfect Square Trinomials<\/h2>\r\nCertain binomial products have special forms. When a binomial is squared, the result is called a <strong>perfect square trinomial<\/strong>. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier. Let\u2019s look at a few perfect square trinomials to familiarize ourselves with the form.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ }{\\left(x+5\\right)}^{2}&amp; =&amp; \\text{ }{x}^{2}+10x+25\\hfill \\\\ \\hfill {\\left(x - 3\\right)}^{2}&amp; =&amp; \\text{ }{x}^{2}-6x+9\\hfill \\\\ \\hfill {\\left(4x - 1\\right)}^{2}&amp; =&amp; 4{x}^{2}-8x+1\\hfill \\end{array}[\/latex]<\/div>\r\nNotice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Perfect Square Trinomials<\/h3>\r\nWhen a binomial is squared, the result is the first term squared added to double the product of both terms and the last term squared.\r\n<div style=\"text-align: center;\">[latex]{\\left(x+a\\right)}^{2}=\\left(x+a\\right)\\left(x+a\\right)={x}^{2}+2ax+{a}^{2}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a binomial, square it using the formula for perfect square trinomials<\/h3>\r\n<ol>\r\n \t<li>Square the first term of the binomial.<\/li>\r\n \t<li>Square the last term of the binomial.<\/li>\r\n \t<li>For the middle term of the trinomial, double the product of the two terms.<\/li>\r\n \t<li>Add and simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Expanding Perfect Squares<\/h3>\r\nExpand [latex]{\\left(3x - 8\\right)}^{2}[\/latex].\r\n\r\n[reveal-answer q=\"733978\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"733978\"]\r\n\r\nBegin by squaring the first term and the last term. For the middle term of the trinomial, double the product of the two terms.\r\n<div style=\"text-align: center;\">[latex]{\\left(3x\\right)}^{2}-2\\left(3x\\right)\\left(8\\right)+{\\left(-8\\right)}^{2}[\/latex]<\/div>\r\n<p style=\"text-align: center;\">[latex]9{x}^{2}-48x+64[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExpand [latex]{\\left(4x - 1\\right)}^{2}[\/latex].\r\n\r\n[reveal-answer q=\"278544\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"278544\"]\r\n\r\n[latex]16{x}^{2}-8x+1[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]1825[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Difference of Squares<\/h2>\r\nAnother special product is called the <strong>difference of squares\u00a0<\/strong>which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let\u2019s see what happens when we multiply [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex] using the FOIL method.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left(x+1\\right)\\left(x - 1\\right)&amp; =&amp; {x}^{2}-x+x - 1\\hfill \\\\ &amp; =&amp; {x}^{2}-1\\hfill \\end{array}[\/latex]<\/div>\r\nThe middle term drops out resulting in a difference of squares. Just as we did with the perfect squares, let\u2019s look at a few examples.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left(x+5\\right)\\left(x - 5\\right)&amp; =&amp; {x}^{2}-25\\hfill \\\\ \\hfill \\left(x+11\\right)\\left(x - 11\\right)&amp; =&amp; {x}^{2}-121\\hfill \\\\ \\hfill \\left(2x+3\\right)\\left(2x - 3\\right)&amp; =&amp; 4{x}^{2}-9\\hfill \\end{array}[\/latex]<\/div>\r\nBecause the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Is there a special form for the sum of squares?<\/strong>\r\n\r\n<em>No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Difference of Squares<\/h3>\r\nWhen a binomial is multiplied by a binomial with the same terms separated by the opposite sign, the result is the square of the first term minus the square of the last term.\r\n<div style=\"text-align: center;\">[latex]\\left(a+b\\right)\\left(a-b\\right)={a}^{2}-{b}^{2}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a binomial multiplied by a binomial with the same terms but the opposite sign, find the difference of squares<\/h3>\r\n<ol>\r\n \t<li>Square the first term of the binomials.<\/li>\r\n \t<li>Square the last term of the binomials.<\/li>\r\n \t<li>Subtract the square of the last term from the square of the first term.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Multiplying Binomials Resulting in a Difference of Squares<\/h3>\r\nMultiply [latex]\\left(9x+4\\right)\\left(9x - 4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"366563\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"366563\"]\r\n\r\nSquare the first term to get [latex]{\\left(9x\\right)}^{2}=81{x}^{2}[\/latex]. Square the last term to get [latex]{4}^{2}=16[\/latex]. Subtract the square of the last term from the square of the first term to find the product of [latex]81{x}^{2}-16[\/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\nMultiply [latex]\\left(2x+7\\right)\\left(2x - 7\\right)[\/latex].\r\n\r\n[reveal-answer q=\"951379\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"951379\"]\r\n\r\n[latex]4{x}^{2}-49[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]1856[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Performing Operations with Polynomials of Several Variables<\/h2>\r\nWe have looked at polynomials containing only one variable. However, a polynomial can contain several variables. All of the same rules apply when working with polynomials containing several variables. Consider an example:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}\\left(a+2b\\right)\\left(4a-b-c\\right)\\hfill &amp; \\hfill \\\\ a\\left(4a-b-c\\right)+2b\\left(4a-b-c\\right)\\hfill &amp; \\text{Use the distributive property}.\\hfill \\\\ 4{a}^{2}-ab-ac+8ab - 2{b}^{2}-2bc\\hfill &amp; \\text{Multiply}.\\hfill \\\\ 4{a}^{2}+\\left(-ab+8ab\\right)-ac - 2{b}^{2}-2bc\\hfill &amp; \\text{Combine like terms}.\\hfill \\\\ 4{a}^{2}+7ab-ac - 2bc - 2{b}^{2}\\hfill &amp; \\text{Simplify}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Multiplying Polynomials Containing Several Variables<\/h3>\r\nMultiply [latex]\\left(x+4\\right)\\left(3x - 2y+5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"313997\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"313997\"]\r\n\r\nFollow the same steps that we used to multiply polynomials containing only one variable.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}x\\left(3x - 2y+5\\right)+4\\left(3x - 2y+5\\right) \\hfill &amp; \\text{Use the distributive property}.\\hfill \\\\ 3{x}^{2}-2xy+5x+12x - 8y+20\\hfill &amp; \\text{Multiply}.\\hfill \\\\ 3{x}^{2}-2xy+\\left(5x+12x\\right)-8y+20\\hfill &amp; \\text{Combine like terms}.\\hfill \\\\ 3{x}^{2}-2xy+17x - 8y+20 \\hfill &amp; \\text{Simplify}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[latex]\\left(3x - 1\\right)\\left(2x+7y - 9\\right)[\/latex].\r\n\r\n[reveal-answer q=\"383366\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"383366\"]\r\n\r\n[latex]6{x}^{2}+21xy - 29x - 7y+9[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]100774[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Square a binomial.<\/li>\n<li>Find a difference of squares.<\/li>\n<li>Perform operations on polynomials with several variables.<\/li>\n<\/ul>\n<\/div>\n<h2>Perfect Square Trinomials<\/h2>\n<p>Certain binomial products have special forms. When a binomial is squared, the result is called a <strong>perfect square trinomial<\/strong>. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier. Let\u2019s look at a few perfect square trinomials to familiarize ourselves with the form.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\text{ }{\\left(x+5\\right)}^{2}& =& \\text{ }{x}^{2}+10x+25\\hfill \\\\ \\hfill {\\left(x - 3\\right)}^{2}& =& \\text{ }{x}^{2}-6x+9\\hfill \\\\ \\hfill {\\left(4x - 1\\right)}^{2}& =& 4{x}^{2}-8x+1\\hfill \\end{array}[\/latex]<\/div>\n<p>Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Perfect Square Trinomials<\/h3>\n<p>When a binomial is squared, the result is the first term squared added to double the product of both terms and the last term squared.<\/p>\n<div style=\"text-align: center;\">[latex]{\\left(x+a\\right)}^{2}=\\left(x+a\\right)\\left(x+a\\right)={x}^{2}+2ax+{a}^{2}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a binomial, square it using the formula for perfect square trinomials<\/h3>\n<ol>\n<li>Square the first term of the binomial.<\/li>\n<li>Square the last term of the binomial.<\/li>\n<li>For the middle term of the trinomial, double the product of the two terms.<\/li>\n<li>Add and simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Expanding Perfect Squares<\/h3>\n<p>Expand [latex]{\\left(3x - 8\\right)}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q733978\">Show Solution<\/span><\/p>\n<div id=\"q733978\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by squaring the first term and the last term. For the middle term of the trinomial, double the product of the two terms.<\/p>\n<div style=\"text-align: center;\">[latex]{\\left(3x\\right)}^{2}-2\\left(3x\\right)\\left(8\\right)+{\\left(-8\\right)}^{2}[\/latex]<\/div>\n<p style=\"text-align: center;\">[latex]9{x}^{2}-48x+64[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Expand [latex]{\\left(4x - 1\\right)}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q278544\">Show Solution<\/span><\/p>\n<div id=\"q278544\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]16{x}^{2}-8x+1[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm1825\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1825&theme=oea&iframe_resize_id=ohm1825&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Difference of Squares<\/h2>\n<p>Another special product is called the <strong>difference of squares\u00a0<\/strong>which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let\u2019s see what happens when we multiply [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex] using the FOIL method.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left(x+1\\right)\\left(x - 1\\right)& =& {x}^{2}-x+x - 1\\hfill \\\\ & =& {x}^{2}-1\\hfill \\end{array}[\/latex]<\/div>\n<p>The middle term drops out resulting in a difference of squares. Just as we did with the perfect squares, let\u2019s look at a few examples.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill \\left(x+5\\right)\\left(x - 5\\right)& =& {x}^{2}-25\\hfill \\\\ \\hfill \\left(x+11\\right)\\left(x - 11\\right)& =& {x}^{2}-121\\hfill \\\\ \\hfill \\left(2x+3\\right)\\left(2x - 3\\right)& =& 4{x}^{2}-9\\hfill \\end{array}[\/latex]<\/div>\n<p>Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.<\/p>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Is there a special form for the sum of squares?<\/strong><\/p>\n<p><em>No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Difference of Squares<\/h3>\n<p>When a binomial is multiplied by a binomial with the same terms separated by the opposite sign, the result is the square of the first term minus the square of the last term.<\/p>\n<div style=\"text-align: center;\">[latex]\\left(a+b\\right)\\left(a-b\\right)={a}^{2}-{b}^{2}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a binomial multiplied by a binomial with the same terms but the opposite sign, find the difference of squares<\/h3>\n<ol>\n<li>Square the first term of the binomials.<\/li>\n<li>Square the last term of the binomials.<\/li>\n<li>Subtract the square of the last term from the square of the first term.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Multiplying Binomials Resulting in a Difference of Squares<\/h3>\n<p>Multiply [latex]\\left(9x+4\\right)\\left(9x - 4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q366563\">Show Solution<\/span><\/p>\n<div id=\"q366563\" class=\"hidden-answer\" style=\"display: none\">\n<p>Square the first term to get [latex]{\\left(9x\\right)}^{2}=81{x}^{2}[\/latex]. Square the last term to get [latex]{4}^{2}=16[\/latex]. Subtract the square of the last term from the square of the first term to find the product of [latex]81{x}^{2}-16[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Multiply [latex]\\left(2x+7\\right)\\left(2x - 7\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q951379\">Show Solution<\/span><\/p>\n<div id=\"q951379\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]4{x}^{2}-49[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm1856\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1856&theme=oea&iframe_resize_id=ohm1856&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Performing Operations with Polynomials of Several Variables<\/h2>\n<p>We have looked at polynomials containing only one variable. However, a polynomial can contain several variables. All of the same rules apply when working with polynomials containing several variables. Consider an example:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}\\left(a+2b\\right)\\left(4a-b-c\\right)\\hfill & \\hfill \\\\ a\\left(4a-b-c\\right)+2b\\left(4a-b-c\\right)\\hfill & \\text{Use the distributive property}.\\hfill \\\\ 4{a}^{2}-ab-ac+8ab - 2{b}^{2}-2bc\\hfill & \\text{Multiply}.\\hfill \\\\ 4{a}^{2}+\\left(-ab+8ab\\right)-ac - 2{b}^{2}-2bc\\hfill & \\text{Combine like terms}.\\hfill \\\\ 4{a}^{2}+7ab-ac - 2bc - 2{b}^{2}\\hfill & \\text{Simplify}.\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Multiplying Polynomials Containing Several Variables<\/h3>\n<p>Multiply [latex]\\left(x+4\\right)\\left(3x - 2y+5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q313997\">Show Solution<\/span><\/p>\n<div id=\"q313997\" class=\"hidden-answer\" style=\"display: none\">\n<p>Follow the same steps that we used to multiply polynomials containing only one variable.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}x\\left(3x - 2y+5\\right)+4\\left(3x - 2y+5\\right) \\hfill & \\text{Use the distributive property}.\\hfill \\\\ 3{x}^{2}-2xy+5x+12x - 8y+20\\hfill & \\text{Multiply}.\\hfill \\\\ 3{x}^{2}-2xy+\\left(5x+12x\\right)-8y+20\\hfill & \\text{Combine like terms}.\\hfill \\\\ 3{x}^{2}-2xy+17x - 8y+20 \\hfill & \\text{Simplify}.\\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>[latex]\\left(3x - 1\\right)\\left(2x+7y - 9\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q383366\">Show Solution<\/span><\/p>\n<div id=\"q383366\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]6{x}^{2}+21xy - 29x - 7y+9[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm100774\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=100774&theme=oea&iframe_resize_id=ohm100774&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-63\">\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>Question ID 1825. <strong>Authored by<\/strong>: David Whittaker. <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 1856. <strong>Authored by<\/strong>: Lawrence Morales. <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 100774. <strong>Authored by<\/strong>: Rick Rieman. <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":9,"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 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