{"id":1807,"date":"2016-11-02T20:49:45","date_gmt":"2016-11-02T20:49:45","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1807"},"modified":"2017-04-19T16:50:29","modified_gmt":"2017-04-19T16:50:29","slug":"polynomial-long-division","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/polynomial-long-division\/","title":{"raw":"Polynomial Long Division","rendered":"Polynomial Long Division"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Use long division to divide a polynomial by a binomial<\/li>\r\n<\/ul>\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\nIn the next two sections, we will be learning two ways to divide polynomials. These techniques can help you find the zeros of a polynomial that\u00a0is not factorable over the integers.\r\n\r\nWe are familiar with the <strong>long division<\/strong> algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let\u2019s divide 178 by 3 using long division.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02204318\/CNX_Precalc_Figure_03_05_0022.jpg\" alt=\"Long Division. Step 1, 5 times 3 equals 15 and 17 minus 15 equals 2. Step 2: Bring down the 8. Step 3: 9 times 3 equals 27 and 28 minus 27 equals 1. Answer: 59 with a remainder of 1 or 59 and one-third.\" width=\"487\" height=\"181\" data-media-type=\"image\/jpg\" \/>\r\n\r\nAnother way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{dividend = }\\left(\\text{divisor }\\cdot \\text{ quotient}\\right)\\text{ + remainder}\\hfill \\\\ 178=\\left(3\\cdot 59\\right)+1\\hfill \\\\ =177+1\\hfill \\\\ =178\\hfill \\end{array}[\/latex]<\/p>\r\nWe call this the <strong>Division Algorithm <\/strong>and will discuss it more formally after looking at an example.\r\n\r\nDivision of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide [latex]2{x}^{3}-3{x}^{2}+4x+5[\/latex]\u00a0by [latex]x+2[\/latex]\u00a0using the long division algorithm, it would look like this:\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02204321\/CNX_Precalc_revised_eq_12.png\" alt=\"Set up the division problem. 2x cubed divided by x is 2x squared. Multiply the sum of x and 2 by 2x squared. Subtract. Then bring down the next term. Negative 7x squared divided by x is negative 7x. Multiply the sum of x and 2 by negative 7x. Subtract, then bring down the next term. 18x divided by x is 18. Multiply the sum of x and 2 by 18. Subtract.\" width=\"522\" height=\"462\" data-media-type=\"image\/jpg\" \/>\r\n\r\nWe have found\r\n<p style=\"text-align: center;\">[latex]\\frac{2{x}^{3}-3{x}^{2}+4x+5}{x+2}=2{x}^{2}-7x+18-\\frac{31}{x+2}[\/latex]<\/p>\r\nor\r\n<p style=\"text-align: center;\">[latex]2{x}^{3}-3{x}^{2}+4x+5=\\left(x+2\\right)\\left(2{x}^{2}-7x+18\\right)-31[\/latex]<\/p>\r\nWe can identify the <strong>dividend<\/strong>, the <strong>divisor<\/strong>, the <strong>quotient<\/strong>, and the <strong>remainder<\/strong>.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02204324\/CNX_Precalc_Figure_03_05_0032.jpg\" alt=\"The dividend is 2x cubed minus 3x squared plus 4x plus 5. The divisor is x plus 2. The quotient is 2x squared minus 7x plus 18. The remainder is negative 31.\" width=\"487\" height=\"99\" data-media-type=\"image\/jpg\" \/>\r\n\r\nWriting the result in this manner illustrates the Division Algorithm.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Division Algorithm<\/h3>\r\nThe <strong>Division Algorithm<\/strong> states that, given a polynomial dividend [latex]f\\left(x\\right)[\/latex]\u00a0and a non-zero polynomial divisor [latex]d\\left(x\\right)[\/latex]\u00a0where the degree of [latex]d\\left(x\\right)[\/latex]\u00a0is less than or equal to the degree of [latex]f\\left(x\\right)[\/latex],\u00a0there exist unique polynomials [latex]q\\left(x\\right)[\/latex]\u00a0and [latex]r\\left(x\\right)[\/latex]\u00a0such that\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=d\\left(x\\right)q\\left(x\\right)+r\\left(x\\right)[\/latex]<\/p>\r\n[latex]q\\left(x\\right)[\/latex]\u00a0is the quotient and [latex]r\\left(x\\right)[\/latex]\u00a0is the remainder. The remainder is either equal to zero or has degree strictly less than [latex]d\\left(x\\right)[\/latex].\r\n\r\nIf [latex]r\\left(x\\right)=0[\/latex],\u00a0then [latex]d\\left(x\\right)[\/latex]\u00a0divides evenly into [latex]f\\left(x\\right)[\/latex].\u00a0This means that, in this case, both [latex]d\\left(x\\right)[\/latex]\u00a0and [latex]q\\left(x\\right)[\/latex]\u00a0are factors of [latex]f\\left(x\\right)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a polynomial and a binomial, use long division to divide the polynomial by the binomial.<\/h3>\r\n<ol>\r\n \t<li>Set up the division problem.<\/li>\r\n \t<li>Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.<\/li>\r\n \t<li>Multiply the answer by the divisor and write it below the like terms of the dividend.<\/li>\r\n \t<li>Subtract the bottom <strong>binomial<\/strong> from the top binomial.<\/li>\r\n \t<li>Bring down the next term of the dividend.<\/li>\r\n \t<li>Repeat steps 2\u20135 until reaching the last term of the dividend.<\/li>\r\n \t<li>If the remainder is non-zero, express as a fraction using the divisor as the denominator.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Long Division to Divide a Second-Degree Polynomial<\/h3>\r\nDivide [latex]5{x}^{2}+3x - 2[\/latex]\u00a0by [latex]x+1[\/latex].\r\n\r\n[reveal-answer q=\"996959\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"996959\"]\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02204327\/CNX_Precalc_revised_eq_22.png\" alt=\"Set up the division problem. 5x squared divided by x is 5x. Multiply x plus 1 by 5x. Subtract. Bring down the next term. Negative 2x divded by x is negative 2. Multiply x + 1 by negative 2. Subtract.\" width=\"426\" height=\"288\" data-media-type=\"image\/jpg\" \/>The quotient is [latex]5x - 2[\/latex].\u00a0The remainder is 0. We write the result as\r\n<p style=\"text-align: center;\">[latex]\\frac{5{x}^{2}+3x - 2}{x+1}=5x - 2[\/latex]<\/p>\r\nor\r\n<p style=\"text-align: center;\">[latex]5{x}^{2}+3x - 2=\\left(x+1\\right)\\left(5x - 2\\right)[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThis division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Long Division to Divide a Third-Degree Polynomial<\/h3>\r\nDivide [latex]6{x}^{3}+11{x}^{2}-31x+15[\/latex]\u00a0by [latex]3x - 2[\/latex].\r\n\r\n[reveal-answer q=\"850001\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"850001\"]\r\n<a href=\"https:\/\/courses.candelalearning.com\/precalcone1xmommaster\/wp-content\/uploads\/sites\/1226\/2015\/08\/replacesquareroot.png\"><img class=\"aligncenter wp-image-11885\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02204330\/replacesquareroot.png\" alt=\"6x cubed divided by 3x is 2x squared. Multiply the sum of x and 2 by 2x squared. Subtract. Bring down the next term. 15x squared divided by 3x is 5x. Multiply 3x minus 2 by 5x. Subtract. Bring down the next term. Negative 21x divided by 3x is negative 7. Multiply 3x minus 2 by negative 7. Subtract. The remainder is 1.\" width=\"621\" height=\"153\" \/><\/a>\r\n\r\nThere is a remainder of 1. We can express the result as:\r\n\r\n[latex]\\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x - 2}=2{x}^{2}+5x - 7+\\frac{1}{3x - 2}[\/latex]\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can check our work by using the Division Algorithm to rewrite the solution. Then multiply.\r\n\r\n[latex]\\left(3x - 2\\right)\\left(2{x}^{2}+5x - 7\\right)+1=6{x}^{3}+11{x}^{2}-31x+15[\/latex]\r\n\r\nNotice, as we write our result,\r\n<ul id=\"fs-id1165135152079\">\r\n \t<li>the dividend is [latex]6{x}^{3}+11{x}^{2}-31x+15[\/latex]<\/li>\r\n \t<li>the divisor is [latex]3x - 2[\/latex]<\/li>\r\n \t<li>the quotient is [latex]2{x}^{2}+5x - 7[\/latex]<\/li>\r\n \t<li>the remainder is\u00a01<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nDivide [latex]16{x}^{3}-12{x}^{2}+20x - 3[\/latex]\u00a0by [latex]4x+5[\/latex].\r\n\r\n[reveal-answer q=\"198989\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"198989\"][latex]4{x}^{2}-8x+15-\\frac{78}{4x+5}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=29482&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Use long division to divide a polynomial by a binomial<\/li>\n<\/ul>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><br \/>\nIn the next two sections, we will be learning two ways to divide polynomials. These techniques can help you find the zeros of a polynomial that\u00a0is not factorable over the integers.<\/p>\n<p>We are familiar with the <strong>long division<\/strong> algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let\u2019s divide 178 by 3 using long division.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02204318\/CNX_Precalc_Figure_03_05_0022.jpg\" alt=\"Long Division. Step 1, 5 times 3 equals 15 and 17 minus 15 equals 2. Step 2: Bring down the 8. Step 3: 9 times 3 equals 27 and 28 minus 27 equals 1. Answer: 59 with a remainder of 1 or 59 and one-third.\" width=\"487\" height=\"181\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{dividend = }\\left(\\text{divisor }\\cdot \\text{ quotient}\\right)\\text{ + remainder}\\hfill \\\\ 178=\\left(3\\cdot 59\\right)+1\\hfill \\\\ =177+1\\hfill \\\\ =178\\hfill \\end{array}[\/latex]<\/p>\n<p>We call this the <strong>Division Algorithm <\/strong>and will discuss it more formally after looking at an example.<\/p>\n<p>Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide [latex]2{x}^{3}-3{x}^{2}+4x+5[\/latex]\u00a0by [latex]x+2[\/latex]\u00a0using the long division algorithm, it would look like this:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02204321\/CNX_Precalc_revised_eq_12.png\" alt=\"Set up the division problem. 2x cubed divided by x is 2x squared. Multiply the sum of x and 2 by 2x squared. Subtract. Then bring down the next term. Negative 7x squared divided by x is negative 7x. Multiply the sum of x and 2 by negative 7x. Subtract, then bring down the next term. 18x divided by x is 18. Multiply the sum of x and 2 by 18. Subtract.\" width=\"522\" height=\"462\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>We have found<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{2{x}^{3}-3{x}^{2}+4x+5}{x+2}=2{x}^{2}-7x+18-\\frac{31}{x+2}[\/latex]<\/p>\n<p>or<\/p>\n<p style=\"text-align: center;\">[latex]2{x}^{3}-3{x}^{2}+4x+5=\\left(x+2\\right)\\left(2{x}^{2}-7x+18\\right)-31[\/latex]<\/p>\n<p>We can identify the <strong>dividend<\/strong>, the <strong>divisor<\/strong>, the <strong>quotient<\/strong>, and the <strong>remainder<\/strong>.<\/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\/11\/02204324\/CNX_Precalc_Figure_03_05_0032.jpg\" alt=\"The dividend is 2x cubed minus 3x squared plus 4x plus 5. The divisor is x plus 2. The quotient is 2x squared minus 7x plus 18. The remainder is negative 31.\" width=\"487\" height=\"99\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>Writing the result in this manner illustrates the Division Algorithm.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Division Algorithm<\/h3>\n<p>The <strong>Division Algorithm<\/strong> states that, given a polynomial dividend [latex]f\\left(x\\right)[\/latex]\u00a0and a non-zero polynomial divisor [latex]d\\left(x\\right)[\/latex]\u00a0where the degree of [latex]d\\left(x\\right)[\/latex]\u00a0is less than or equal to the degree of [latex]f\\left(x\\right)[\/latex],\u00a0there exist unique polynomials [latex]q\\left(x\\right)[\/latex]\u00a0and [latex]r\\left(x\\right)[\/latex]\u00a0such that<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=d\\left(x\\right)q\\left(x\\right)+r\\left(x\\right)[\/latex]<\/p>\n<p>[latex]q\\left(x\\right)[\/latex]\u00a0is the quotient and [latex]r\\left(x\\right)[\/latex]\u00a0is the remainder. The remainder is either equal to zero or has degree strictly less than [latex]d\\left(x\\right)[\/latex].<\/p>\n<p>If [latex]r\\left(x\\right)=0[\/latex],\u00a0then [latex]d\\left(x\\right)[\/latex]\u00a0divides evenly into [latex]f\\left(x\\right)[\/latex].\u00a0This means that, in this case, both [latex]d\\left(x\\right)[\/latex]\u00a0and [latex]q\\left(x\\right)[\/latex]\u00a0are factors of [latex]f\\left(x\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a polynomial and a binomial, use long division to divide the polynomial by the binomial.<\/h3>\n<ol>\n<li>Set up the division problem.<\/li>\n<li>Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.<\/li>\n<li>Multiply the answer by the divisor and write it below the like terms of the dividend.<\/li>\n<li>Subtract the bottom <strong>binomial<\/strong> from the top binomial.<\/li>\n<li>Bring down the next term of the dividend.<\/li>\n<li>Repeat steps 2\u20135 until reaching the last term of the dividend.<\/li>\n<li>If the remainder is non-zero, express as a fraction using the divisor as the denominator.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Long Division to Divide a Second-Degree Polynomial<\/h3>\n<p>Divide [latex]5{x}^{2}+3x - 2[\/latex]\u00a0by [latex]x+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q996959\">Solution<\/span><\/p>\n<div id=\"q996959\" class=\"hidden-answer\" style=\"display: none\">\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02204327\/CNX_Precalc_revised_eq_22.png\" alt=\"Set up the division problem. 5x squared divided by x is 5x. Multiply x plus 1 by 5x. Subtract. Bring down the next term. Negative 2x divded by x is negative 2. Multiply x + 1 by negative 2. Subtract.\" width=\"426\" height=\"288\" data-media-type=\"image\/jpg\" \/>The quotient is [latex]5x - 2[\/latex].\u00a0The remainder is 0. We write the result as<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{5{x}^{2}+3x - 2}{x+1}=5x - 2[\/latex]<\/p>\n<p>or<\/p>\n<p style=\"text-align: center;\">[latex]5{x}^{2}+3x - 2=\\left(x+1\\right)\\left(5x - 2\\right)[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Long Division to Divide a Third-Degree Polynomial<\/h3>\n<p>Divide [latex]6{x}^{3}+11{x}^{2}-31x+15[\/latex]\u00a0by [latex]3x - 2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q850001\">Solution<\/span><\/p>\n<div id=\"q850001\" class=\"hidden-answer\" style=\"display: none\">\n<a href=\"https:\/\/courses.candelalearning.com\/precalcone1xmommaster\/wp-content\/uploads\/sites\/1226\/2015\/08\/replacesquareroot.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-11885\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02204330\/replacesquareroot.png\" alt=\"6x cubed divided by 3x is 2x squared. Multiply the sum of x and 2 by 2x squared. Subtract. Bring down the next term. 15x squared divided by 3x is 5x. Multiply 3x minus 2 by 5x. Subtract. Bring down the next term. Negative 21x divided by 3x is negative 7. Multiply 3x minus 2 by negative 7. Subtract. The remainder is 1.\" width=\"621\" height=\"153\" \/><\/a><\/p>\n<p>There is a remainder of 1. We can express the result as:<\/p>\n<p>[latex]\\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x - 2}=2{x}^{2}+5x - 7+\\frac{1}{3x - 2}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can check our work by using the Division Algorithm to rewrite the solution. Then multiply.<\/p>\n<p>[latex]\\left(3x - 2\\right)\\left(2{x}^{2}+5x - 7\\right)+1=6{x}^{3}+11{x}^{2}-31x+15[\/latex]<\/p>\n<p>Notice, as we write our result,<\/p>\n<ul id=\"fs-id1165135152079\">\n<li>the dividend is [latex]6{x}^{3}+11{x}^{2}-31x+15[\/latex]<\/li>\n<li>the divisor is [latex]3x - 2[\/latex]<\/li>\n<li>the quotient is [latex]2{x}^{2}+5x - 7[\/latex]<\/li>\n<li>the remainder is\u00a01<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Divide [latex]16{x}^{3}-12{x}^{2}+20x - 3[\/latex]\u00a0by [latex]4x+5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q198989\">Solution<\/span><\/p>\n<div id=\"q198989\" class=\"hidden-answer\" style=\"display: none\">[latex]4{x}^{2}-8x+15-\\frac{78}{4x+5}[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=29482&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/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-1807\">\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>Question ID 29482. <strong>Authored by<\/strong>: McClure,Caren. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><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><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Question ID 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