{"id":219,"date":"2023-06-21T13:22:45","date_gmt":"2023-06-21T13:22:45","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/polynomial-long-division\/"},"modified":"2023-07-04T03:59:23","modified_gmt":"2023-07-04T03:59:23","slug":"polynomial-long-division","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/polynomial-long-division\/","title":{"raw":"\u25aa   Polynomial Long Division","rendered":"\u25aa   Polynomial Long Division"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use long division to divide polynomials.<\/li>\r\n<\/ul>\r\n<\/div>\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<div class=\"textbox examples\">\r\n<h3>recall long division<\/h3>\r\nA handy algebraic technique involves rewriting a quotient of polynomials as a polynomial of reduced degree plus a remainder quotient\r\n<p style=\"text-align: center;\">[latex]\\dfrac{p(x)}{q(x)}=h(x)+\\dfrac{r(x)}{q(x)}[\/latex].<\/p>\r\nOr equivalently by multiplying both sides by [latex]q(x)[\/latex],\r\n<p style=\"text-align: center;\">[latex]p(x)=q(x)h(x) + r(x)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">This method is akin to rewriting an \"improper\" fraction such as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{17}{3} = 5 + \\dfrac{2}{3}[\/latex].<\/p>\r\nOr equivalently,\r\n<p style=\"text-align: center;\">[latex]17=3\\cdot 5 + 2[\/latex].<\/p>\r\n<p style=\"text-align: left;\">The process used to accomplish this is similar to long division of whole numbers, so if you haven't performed long division for awhile, take a moment to refresh before reading this section.<\/p>\r\n[reveal-answer q=\"914465\"]more[\/reveal-answer]\r\n[hidden-answer a=\"914465\"]\r\n\r\nRecall how you can use long division to divide two whole numbers, say\u00a0[latex]900[\/latex] divided by\u00a0[latex]37[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152919\/image007.jpg\" alt=\"The dividend in 900 and the divisor is 37.\" width=\"51\" height=\"18\" \/>\r\n\r\nFirst, you would think about how many\u00a0[latex]37s[\/latex] are in\u00a0[latex]90[\/latex], as\u00a0[latex]9[\/latex] is too small. (<i>Note: <\/i>you could also think, how many\u00a0[latex]40s[\/latex] are there in\u00a0[latex]90[\/latex].)\r\n\r\nThere are two\u00a0[latex]37s[\/latex] in\u00a0[latex]90[\/latex], so write\u00a0[latex]2[\/latex] above the last digit of\u00a0[latex]90[\/latex]. Two\u00a0[latex]37s[\/latex] is\u00a0[latex]74[\/latex]; write that product below the\u00a0[latex]90[\/latex].\r\n\r\n<img class=\"aligncenter wp-image-2251 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152920\/Screen-Shot-2016-03-28-at-3.35.17-PM.png\" alt=\"Screen Shot 2016-03-28 at 3.35.17 PM\" width=\"69\" height=\"55\" \/>\r\nSubtract: [latex]90\u201374[\/latex] is\u00a0[latex]16[\/latex]. (If the result is larger than the divisor,\u00a0[latex]37[\/latex], then you need to use a larger number for the quotient.)\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152921\/image008.jpg\" alt=\"\" width=\"51\" height=\"57\" \/>\r\nBring down the next digit\u00a0[latex](0)[\/latex] and consider how many\u00a0[latex]37s[\/latex] are in\u00a0[latex]160[\/latex].\r\n\r\n<img class=\"aligncenter size-full wp-image-2252\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152922\/Screen-Shot-2016-03-28-at-3.36.06-PM.png\" alt=\"Screen Shot 2016-03-28 at 3.36.06 PM\" width=\"66\" height=\"68\" \/>\r\n\r\nThere are four\u00a0[latex]37s[\/latex] in\u00a0[latex]160[\/latex], so write the\u00a0[latex]4[\/latex] next to the two in the quotient. Four\u00a0[latex]37s[\/latex] is\u00a0[latex]148[\/latex]; write that product below the\u00a0[latex]160[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152923\/image009.jpg\" alt=\"\" width=\"51\" height=\"74\" \/>\r\nSubtract: [latex]160\u2013148[\/latex] is\u00a0[latex]12[\/latex]. This is less than\u00a0[latex]37[\/latex] so the\u00a0[latex]4[\/latex] is correct. Since there are no more digits in the dividend to bring down, you are done.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152924\/image010.jpg\" alt=\"\" width=\"51\" height=\"86\" \/>\r\n\r\nThe final answer is\u00a0[latex]24[\/latex] R[latex]12[\/latex], or [latex]24\\frac{12}{37}[\/latex]. You can check this by multiplying the quotient (without the remainder) by the divisor, and then adding in the remainder. The result should be the dividend:\r\n<p style=\"text-align: center;\">[latex]24\\cdot37+12=888+12=900[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\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\" \/>\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}\\left(\\text{divisor }\\cdot \\text{ quotient}\\right)\\text{ + remainder}\\text{ = dividend}\\hfill \\\\ \\left(3\\cdot 59\\right)+1 = 177+1 = 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\r\n<img class=\"aligncenter size-full wp-image-995\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4695\/2019\/07\/25211416\/Screenshot_20230125_0411441.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=\"617\" height=\"609\" \/>\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\n<p style=\"text-align: center;\">or<\/p>\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>,\u00a0<strong>divisor<\/strong>,\u00a0<strong>quotient<\/strong>, and\u00a0<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\" \/>\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 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 terms above it.<\/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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"996959\"]\r\n\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\" \/>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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"850001\"]\r\n\r\n<img class=\"aligncenter size-full wp-image-997\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4695\/2019\/07\/25211535\/Screenshot_20230125_041500.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=\"716\" height=\"156\" \/>\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 multiplying.\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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"198989\"]\r\n\r\n[latex]4{x}^{2}-8x+15-\\frac{78}{4x+5}[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]29482[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use long division to divide polynomials.<\/li>\n<\/ul>\n<\/div>\n<p>In 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<div class=\"textbox examples\">\n<h3>recall long division<\/h3>\n<p>A handy algebraic technique involves rewriting a quotient of polynomials as a polynomial of reduced degree plus a remainder quotient<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{p(x)}{q(x)}=h(x)+\\dfrac{r(x)}{q(x)}[\/latex].<\/p>\n<p>Or equivalently by multiplying both sides by [latex]q(x)[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]p(x)=q(x)h(x) + r(x)[\/latex]<\/p>\n<p style=\"text-align: left;\">This method is akin to rewriting an &#8220;improper&#8221; fraction such as<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{17}{3} = 5 + \\dfrac{2}{3}[\/latex].<\/p>\n<p>Or equivalently,<\/p>\n<p style=\"text-align: center;\">[latex]17=3\\cdot 5 + 2[\/latex].<\/p>\n<p style=\"text-align: left;\">The process used to accomplish this is similar to long division of whole numbers, so if you haven&#8217;t performed long division for awhile, take a moment to refresh before reading this section.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q914465\">more<\/span><\/p>\n<div id=\"q914465\" class=\"hidden-answer\" style=\"display: none\">\n<p>Recall how you can use long division to divide two whole numbers, say\u00a0[latex]900[\/latex] divided by\u00a0[latex]37[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152919\/image007.jpg\" alt=\"The dividend in 900 and the divisor is 37.\" width=\"51\" height=\"18\" \/><\/p>\n<p>First, you would think about how many\u00a0[latex]37s[\/latex] are in\u00a0[latex]90[\/latex], as\u00a0[latex]9[\/latex] is too small. (<i>Note: <\/i>you could also think, how many\u00a0[latex]40s[\/latex] are there in\u00a0[latex]90[\/latex].)<\/p>\n<p>There are two\u00a0[latex]37s[\/latex] in\u00a0[latex]90[\/latex], so write\u00a0[latex]2[\/latex] above the last digit of\u00a0[latex]90[\/latex]. Two\u00a0[latex]37s[\/latex] is\u00a0[latex]74[\/latex]; write that product below the\u00a0[latex]90[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2251 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152920\/Screen-Shot-2016-03-28-at-3.35.17-PM.png\" alt=\"Screen Shot 2016-03-28 at 3.35.17 PM\" width=\"69\" height=\"55\" \/><br \/>\nSubtract: [latex]90\u201374[\/latex] is\u00a0[latex]16[\/latex]. (If the result is larger than the divisor,\u00a0[latex]37[\/latex], then you need to use a larger number for the quotient.)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152921\/image008.jpg\" alt=\"\" width=\"51\" height=\"57\" \/><br \/>\nBring down the next digit\u00a0[latex](0)[\/latex] and consider how many\u00a0[latex]37s[\/latex] are in\u00a0[latex]160[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2252\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152922\/Screen-Shot-2016-03-28-at-3.36.06-PM.png\" alt=\"Screen Shot 2016-03-28 at 3.36.06 PM\" width=\"66\" height=\"68\" \/><\/p>\n<p>There are four\u00a0[latex]37s[\/latex] in\u00a0[latex]160[\/latex], so write the\u00a0[latex]4[\/latex] next to the two in the quotient. Four\u00a0[latex]37s[\/latex] is\u00a0[latex]148[\/latex]; write that product below the\u00a0[latex]160[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152923\/image009.jpg\" alt=\"\" width=\"51\" height=\"74\" \/><br \/>\nSubtract: [latex]160\u2013148[\/latex] is\u00a0[latex]12[\/latex]. This is less than\u00a0[latex]37[\/latex] so the\u00a0[latex]4[\/latex] is correct. Since there are no more digits in the dividend to bring down, you are done.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152924\/image010.jpg\" alt=\"\" width=\"51\" height=\"86\" \/><\/p>\n<p>The final answer is\u00a0[latex]24[\/latex] R[latex]12[\/latex], or [latex]24\\frac{12}{37}[\/latex]. You can check this by multiplying the quotient (without the remainder) by the divisor, and then adding in the remainder. The result should be the dividend:<\/p>\n<p style=\"text-align: center;\">[latex]24\\cdot37+12=888+12=900[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>=======================================================================================<br \/>\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.<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\" \/><\/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}\\left(\\text{divisor }\\cdot \\text{ quotient}\\right)\\text{ + remainder}\\text{ = dividend}\\hfill \\\\ \\left(3\\cdot 59\\right)+1 = 177+1 = 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:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-995\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4695\/2019\/07\/25211416\/Screenshot_20230125_0411441.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=\"617\" height=\"609\" \/><\/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 style=\"text-align: center;\">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>,\u00a0<strong>divisor<\/strong>,\u00a0<strong>quotient<\/strong>, and\u00a0<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\" \/><\/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 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 terms above it.<\/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\">Show Solution<\/span><\/p>\n<div id=\"q996959\" class=\"hidden-answer\" style=\"display: none\">\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\/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\" \/>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\">Show Solution<\/span><\/p>\n<div id=\"q850001\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-997\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4695\/2019\/07\/25211535\/Screenshot_20230125_041500.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=\"716\" height=\"156\" \/><\/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 multiplying.<\/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\">Show Solution<\/span><\/p>\n<div id=\"q198989\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]4{x}^{2}-8x+15-\\frac{78}{4x+5}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm29482\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29482&theme=oea&iframe_resize_id=ohm29482&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-219\">\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><li>Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program.. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext.%20\">http:\/\/nrocnetwork.org\/dm-opentext.%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><\/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":16,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Question ID 29482\",\"author\":\"McClure,Caren\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"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\":\"Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program.\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext. \",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"cf4ca3f5-16e9-4873-8b35-5af286fb0958","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-219","chapter","type-chapter","status-publish","hentry"],"part":202,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/219","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\/219\/revisions"}],"predecessor-version":[{"id":776,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/219\/revisions\/776"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/202"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/219\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=219"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=219"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=219"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=219"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}