{"id":2590,"date":"2016-07-15T17:14:18","date_gmt":"2016-07-15T17:14:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2590"},"modified":"2019-08-06T18:43:59","modified_gmt":"2019-08-06T18:43:59","slug":"read-divide-polynomials-part-i","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-divide-polynomials-part-i\/","title":{"raw":"Divide Polynomials Part I","rendered":"Divide Polynomials Part I"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcome<\/h3>\r\n<ul>\r\n \t<li>Multiply and divide polynomials<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Divide a Polynomial by a Binomial<\/h2>\r\nDividing a <strong>polynomial<\/strong> by a monomial can be handled by dividing each term in the polynomial separately. This cannot be done when the divisor has more than one term. However, the process of long division can be very helpful with polynomials.\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\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\nTo divide polynomials, use the same process. This example shows how to do this when dividing by a <strong>binomial<\/strong>.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nDivide:\u00a0[latex]\\frac{\\left(x^{2}\u20134x\u201312\\right)}{\\left(x+2\\right)}[\/latex]\r\n[reveal-answer q=\"455187\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455187\"]\r\n\r\nHow many <i>x<\/i>\u2019s are there in [latex]x^{2}[\/latex]? That is, what is [latex] \\frac{{{x}^{2}}}{x}[\/latex]?\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152924\/image011.jpg\" alt=\"\" width=\"118\" height=\"18\" \/>\r\n\r\n[latex] \\frac{{{x}^{2}}}{x}=x[\/latex]<i>. <\/i>Put <i>x<\/i> in the quotient above the [latex]-4x[\/latex]<i>\u00a0<\/i>term. These are like terms, which helps to organize the problem.\r\n\r\nWrite the product of the divisor and the part of the quotient you just found under the dividend. Since [latex]x\\left(x+2\\right)=x^{2}+2x[\/latex],\u00a0write this underneath, and get ready to subtract.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152925\/image012.jpg\" alt=\"\" width=\"118\" height=\"51\" \/>\r\n\r\nRewrite [latex]\u2013\\left(x^{2} + 2x\\right)[\/latex]\u00a0as its opposite [latex]\u2013x^{2}\u20132x[\/latex]\u00a0so that you can add the opposite. Adding the opposite is the same as subtracting, and it is easier to do.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152926\/image013.jpg\" alt=\"\" width=\"118\" height=\"48\" \/>\r\n\r\nAdd\u00a0[latex]-x^{2}[\/latex] to [latex]x^{2}[\/latex] and [latex]-2x[\/latex] to [latex]-4x[\/latex].\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152927\/image014.jpg\" alt=\"\" width=\"118\" height=\"60\" \/>\r\n\r\nBring down [latex]-12[\/latex].\r\n\r\n<img class=\"aligncenter size-full wp-image-2253\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152928\/Screen-Shot-2016-03-28-at-4.19.50-PM.png\" alt=\"Screen Shot 2016-03-28 at 4.19.50 PM\" width=\"137\" height=\"75\" \/>\r\n\r\nRepeat the process. How many times does <i>x<\/i> go into [latex]-6x[\/latex]? In other words, what is [latex] \\frac{-6x}{x}[\/latex]?\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152929\/image015.jpg\" alt=\"\" width=\"118\" height=\"60\" \/>\r\n\r\nSince [latex] \\frac{-6x}{x}=-6[\/latex], write [latex]-6[\/latex] in the quotient. Multiply [latex]-6[\/latex] and [latex]x+2[\/latex]\u00a0and prepare to subtract the product.\r\n\r\n<img class=\"aligncenter size-full wp-image-2254\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152931\/Screen-Shot-2016-03-28-at-4.24.52-PM.png\" alt=\"Screen Shot 2016-03-28 at 4.24.52 PM\" width=\"165\" height=\"113\" \/>\r\n\r\nRewrite [latex]\u2013\\left(-6x\u201312\\right)[\/latex] as [latex]6x+12[\/latex], so that you can add the opposite.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152932\/image016.jpg\" alt=\"\" width=\"118\" height=\"80\" \/>\r\n\r\nAdd. In this case, there is no remainder, so you are done.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152933\/image017.jpg\" alt=\"\" width=\"118\" height=\"90\" \/>\r\n\r\nThe answer is [latex]x\u20136[\/latex].\r\n\r\nCheck this by multiplying:\r\n<p style=\"text-align: left;\">[latex]\\left(x-6\\right)\\left(x+2\\right)=x^{2}+2x-6x-12=x^{2}-4x-12[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of dividing a degree two trinomial by a degree one binomial.\r\n\r\nhttps:\/\/youtu.be\/KUPFg__Djzw\r\n\r\nLet us try another example. In this example, a term is \u201cmissing\u201d from the dividend.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nDivide: [latex]\\frac{\\left(x^{3}\u20136x\u201310\\right)}{\\left(x\u20133\\right)}[\/latex]\r\n[reveal-answer q=\"523374\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"523374\"]\r\n\r\nIn setting up this problem, notice that there is an [latex]x^{3}[\/latex]\u00a0term but no [latex]x^{2}[\/latex]\u00a0term. Add [latex]0x^{2}[\/latex]\u00a0as a \u201cplace holder\u201d for this term. Since 0 times anything is 0, you are not changing the value of the dividend.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152934\/image018.jpg\" alt=\"\" width=\"153\" height=\"18\" \/>\r\n\r\nFocus on the first terms again: how many <i>x<\/i>\u2019s are there in [latex]x^{3}[\/latex]? Since [latex] \\frac{{{x}^{3}}}{x}=x^{2}[\/latex], put [latex]x^{2}[\/latex]\u00a0in the quotient.\r\n\r\nMultiply: [latex]x^{2}\\left(x\u20133\\right)=x^{3}\u20133x^{2}[\/latex]; write this underneath the dividend, and prepare to subtract.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152935\/image019.jpg\" alt=\"\" width=\"153\" height=\"49\" \/>\r\n\r\nRewrite the subtraction using the opposite of the expression [latex]x^{3}-3x^{2}[\/latex]. Then add.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152936\/image020.jpg\" alt=\"\" width=\"153\" height=\"59\" \/>\r\n\r\nBring down the rest of the expression in the dividend. It is helpful to bring down <i>all<\/i> of the remaining terms.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152938\/image021.jpg\" alt=\"\" width=\"153\" height=\"59\" \/>\r\n\r\nNow, repeat the process with the remaining expression, [latex]3x^{2}-6x\u201310[\/latex], as the dividend.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152939\/image022.jpg\" alt=\"\" width=\"153\" height=\"79\" \/>\r\n\r\nRemember to watch the signs!\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152940\/image023.jpg\" alt=\"\" width=\"153\" height=\"89\" \/>\r\n\r\nHow many <i>x<\/i>\u2019s are there in\u00a0[latex]3[\/latex]<i>x<\/i>? Since there are\u00a0[latex]3[\/latex], multiply [latex]3\\left(x\u20133\\right)=3x\u20139[\/latex], write this underneath the dividend, and prepare to subtract.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152941\/image024.jpg\" alt=\"\" width=\"157\" height=\"118\" \/>\r\n\r\nContinue until the <strong>degree<\/strong> of the remainder is <i>less <\/i>than the degree of the divisor. In this case the degree of the remainder, [latex]-1[\/latex], is\u00a0[latex]0[\/latex], which is less than the degree of [latex]x-3[\/latex], which is\u00a0[latex]1[\/latex].\r\n\r\nAlso notice that you have brought down all the terms in the dividend and that the quotient extends to the right edge of the dividend. These are other ways to check whether you have completed the problem.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152942\/image025.jpg\" alt=\"\" width=\"153\" height=\"118\" \/>\r\nYou can write the remainder using the symbol R or as a fraction added to the rest of the quotient with the remainder in the numerator and the divisor in the denominator. In this case, since the remainder is negative, you can also subtract the opposite.\r\n\r\nPossible forms of the solution are:\r\n\r\n[latex]\\begin{array}{l}x^{2}+3x+3+R-1,\\\\x^{2}+3x+3+\\frac{-1}{x-3}, \\text{ or }\\\\x^{2}+3x+3-\\frac{1}{x-3}\\end{array}[\/latex]\r\n\r\nCheck the result:\r\n\r\n[latex]\\begin{array}{l}\\left(x\u20133\\right)\\left(x^{2}+3x+3\\right) &amp; =x\\left(x^{2}+3x+3\\right)\u20133\\left(x^{2}+3x+3\\right)\\\\ &amp; =x^{3}+3x^{2}+3x\u20133x^{2}\u20139x\u20139\\\\ &amp; =x^{3}\u20136x\u20139\\end{array}[\/latex]\r\n\r\n[latex]x^{3}\u20136x\u20139+\\left(-1\\right)=x^{3}\u20136x\u201310[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, we show another example of dividing a degree three trinomial by a binomial. Note the \"missing\" term and how we work with it.\r\n\r\nhttps:\/\/youtu.be\/Rxds7Q_UTeo\r\n\r\nThe process above works for dividing any polynomials no matter how many terms are in the divisor or the dividend. The main things to remember are:\r\n<ul>\r\n \t<li>When subtracting, be sure to subtract the whole expression, not just the first term. <i>This is very easy to forget, so be careful!<\/i><\/li>\r\n \t<li>Stop when the degree of the remainder is less than the degree of the dividend or when you have brought down all the terms in the dividend and your quotient extends to the right edge of the dividend.<\/li>\r\n<\/ul>\r\nIn the next video, we present one more example of polynomial long division.\r\n\r\nhttps:\/\/youtu.be\/P6OTbUf8f60\r\n<h2>Summary<\/h2>\r\nDividing polynomials by polynomials of more than one term can be done using a process very much like long division of whole numbers. You must be careful to subtract entire expressions, not just the first term. Stop when the degree of the remainder is less than the degree of the divisor. The remainder can be written using R notation or as a fraction added to the quotient with the remainder in the numerator and the divisor in the denominator.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcome<\/h3>\n<ul>\n<li>Multiply and divide polynomials<\/li>\n<\/ul>\n<\/div>\n<h2>Divide a Polynomial by a Binomial<\/h2>\n<p>Dividing a <strong>polynomial<\/strong> by a monomial can be handled by dividing each term in the polynomial separately. This cannot be done when the divisor has more than one term. However, the process of long division can be very helpful with polynomials.<\/p>\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\" \/><br \/>\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].)<\/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<p>To divide polynomials, use the same process. This example shows how to do this when dividing by a <strong>binomial<\/strong>.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Divide:\u00a0[latex]\\frac{\\left(x^{2}\u20134x\u201312\\right)}{\\left(x+2\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455187\">Show Solution<\/span><\/p>\n<div id=\"q455187\" class=\"hidden-answer\" style=\"display: none\">\n<p>How many <i>x<\/i>\u2019s are there in [latex]x^{2}[\/latex]? That is, what is [latex]\\frac{{{x}^{2}}}{x}[\/latex]?<br \/>\n<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\/image011.jpg\" alt=\"\" width=\"118\" height=\"18\" \/><\/p>\n<p>[latex]\\frac{{{x}^{2}}}{x}=x[\/latex]<i>. <\/i>Put <i>x<\/i> in the quotient above the [latex]-4x[\/latex]<i>\u00a0<\/i>term. These are like terms, which helps to organize the problem.<\/p>\n<p>Write the product of the divisor and the part of the quotient you just found under the dividend. Since [latex]x\\left(x+2\\right)=x^{2}+2x[\/latex],\u00a0write this underneath, and get ready to subtract.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152925\/image012.jpg\" alt=\"\" width=\"118\" height=\"51\" \/><\/p>\n<p>Rewrite [latex]\u2013\\left(x^{2} + 2x\\right)[\/latex]\u00a0as its opposite [latex]\u2013x^{2}\u20132x[\/latex]\u00a0so that you can add the opposite. Adding the opposite is the same as subtracting, and it is easier to do.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152926\/image013.jpg\" alt=\"\" width=\"118\" height=\"48\" \/><\/p>\n<p>Add\u00a0[latex]-x^{2}[\/latex] to [latex]x^{2}[\/latex] and [latex]-2x[\/latex] to [latex]-4x[\/latex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152927\/image014.jpg\" alt=\"\" width=\"118\" height=\"60\" \/><\/p>\n<p>Bring down [latex]-12[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2253\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152928\/Screen-Shot-2016-03-28-at-4.19.50-PM.png\" alt=\"Screen Shot 2016-03-28 at 4.19.50 PM\" width=\"137\" height=\"75\" \/><\/p>\n<p>Repeat the process. How many times does <i>x<\/i> go into [latex]-6x[\/latex]? In other words, what is [latex]\\frac{-6x}{x}[\/latex]?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152929\/image015.jpg\" alt=\"\" width=\"118\" height=\"60\" \/><\/p>\n<p>Since [latex]\\frac{-6x}{x}=-6[\/latex], write [latex]-6[\/latex] in the quotient. Multiply [latex]-6[\/latex] and [latex]x+2[\/latex]\u00a0and prepare to subtract the product.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2254\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152931\/Screen-Shot-2016-03-28-at-4.24.52-PM.png\" alt=\"Screen Shot 2016-03-28 at 4.24.52 PM\" width=\"165\" height=\"113\" \/><\/p>\n<p>Rewrite [latex]\u2013\\left(-6x\u201312\\right)[\/latex] as [latex]6x+12[\/latex], so that you can add the opposite.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152932\/image016.jpg\" alt=\"\" width=\"118\" height=\"80\" \/><\/p>\n<p>Add. In this case, there is no remainder, so you are done.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152933\/image017.jpg\" alt=\"\" width=\"118\" height=\"90\" \/><\/p>\n<p>The answer is [latex]x\u20136[\/latex].<\/p>\n<p>Check this by multiplying:<\/p>\n<p style=\"text-align: left;\">[latex]\\left(x-6\\right)\\left(x+2\\right)=x^{2}+2x-6x-12=x^{2}-4x-12[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of dividing a degree two trinomial by a degree one binomial.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Divide a Trinomial by a Binomial Using Long Division\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/KUPFg__Djzw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Let us try another example. In this example, a term is \u201cmissing\u201d from the dividend.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Divide: [latex]\\frac{\\left(x^{3}\u20136x\u201310\\right)}{\\left(x\u20133\\right)}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q523374\">Show Solution<\/span><\/p>\n<div id=\"q523374\" class=\"hidden-answer\" style=\"display: none\">\n<p>In setting up this problem, notice that there is an [latex]x^{3}[\/latex]\u00a0term but no [latex]x^{2}[\/latex]\u00a0term. Add [latex]0x^{2}[\/latex]\u00a0as a \u201cplace holder\u201d for this term. Since 0 times anything is 0, you are not changing the value of the dividend.<\/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\/15152934\/image018.jpg\" alt=\"\" width=\"153\" height=\"18\" \/><\/p>\n<p>Focus on the first terms again: how many <i>x<\/i>\u2019s are there in [latex]x^{3}[\/latex]? Since [latex]\\frac{{{x}^{3}}}{x}=x^{2}[\/latex], put [latex]x^{2}[\/latex]\u00a0in the quotient.<\/p>\n<p>Multiply: [latex]x^{2}\\left(x\u20133\\right)=x^{3}\u20133x^{2}[\/latex]; write this underneath the dividend, and prepare to subtract.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152935\/image019.jpg\" alt=\"\" width=\"153\" height=\"49\" \/><\/p>\n<p>Rewrite the subtraction using the opposite of the expression [latex]x^{3}-3x^{2}[\/latex]. Then add.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152936\/image020.jpg\" alt=\"\" width=\"153\" height=\"59\" \/><\/p>\n<p>Bring down the rest of the expression in the dividend. It is helpful to bring down <i>all<\/i> of the remaining terms.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152938\/image021.jpg\" alt=\"\" width=\"153\" height=\"59\" \/><\/p>\n<p>Now, repeat the process with the remaining expression, [latex]3x^{2}-6x\u201310[\/latex], as the dividend.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152939\/image022.jpg\" alt=\"\" width=\"153\" height=\"79\" \/><\/p>\n<p>Remember to watch the signs!<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152940\/image023.jpg\" alt=\"\" width=\"153\" height=\"89\" \/><\/p>\n<p>How many <i>x<\/i>\u2019s are there in\u00a0[latex]3[\/latex]<i>x<\/i>? Since there are\u00a0[latex]3[\/latex], multiply [latex]3\\left(x\u20133\\right)=3x\u20139[\/latex], write this underneath the dividend, and prepare to subtract.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152941\/image024.jpg\" alt=\"\" width=\"157\" height=\"118\" \/><\/p>\n<p>Continue until the <strong>degree<\/strong> of the remainder is <i>less <\/i>than the degree of the divisor. In this case the degree of the remainder, [latex]-1[\/latex], is\u00a0[latex]0[\/latex], which is less than the degree of [latex]x-3[\/latex], which is\u00a0[latex]1[\/latex].<\/p>\n<p>Also notice that you have brought down all the terms in the dividend and that the quotient extends to the right edge of the dividend. These are other ways to check whether you have completed the problem.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/15152942\/image025.jpg\" alt=\"\" width=\"153\" height=\"118\" \/><br \/>\nYou can write the remainder using the symbol R or as a fraction added to the rest of the quotient with the remainder in the numerator and the divisor in the denominator. In this case, since the remainder is negative, you can also subtract the opposite.<\/p>\n<p>Possible forms of the solution are:<\/p>\n<p>[latex]\\begin{array}{l}x^{2}+3x+3+R-1,\\\\x^{2}+3x+3+\\frac{-1}{x-3}, \\text{ or }\\\\x^{2}+3x+3-\\frac{1}{x-3}\\end{array}[\/latex]<\/p>\n<p>Check the result:<\/p>\n<p>[latex]\\begin{array}{l}\\left(x\u20133\\right)\\left(x^{2}+3x+3\\right) & =x\\left(x^{2}+3x+3\\right)\u20133\\left(x^{2}+3x+3\\right)\\\\ & =x^{3}+3x^{2}+3x\u20133x^{2}\u20139x\u20139\\\\ & =x^{3}\u20136x\u20139\\end{array}[\/latex]<\/p>\n<p>[latex]x^{3}\u20136x\u20139+\\left(-1\\right)=x^{3}\u20136x\u201310[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video that follows, we show another example of dividing a degree three trinomial by a binomial. Note the &#8220;missing&#8221; term and how we work with it.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Divide a Degree 3 Polynomial by a Degree 1 Polynomial (Long Division with Missing Term)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Rxds7Q_UTeo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The process above works for dividing any polynomials no matter how many terms are in the divisor or the dividend. The main things to remember are:<\/p>\n<ul>\n<li>When subtracting, be sure to subtract the whole expression, not just the first term. <i>This is very easy to forget, so be careful!<\/i><\/li>\n<li>Stop when the degree of the remainder is less than the degree of the dividend or when you have brought down all the terms in the dividend and your quotient extends to the right edge of the dividend.<\/li>\n<\/ul>\n<p>In the next video, we present one more example of polynomial long division.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 6:  Divide a Polynomial by a Degree Two Binomial Using Long Division\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/P6OTbUf8f60?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>Dividing polynomials by polynomials of more than one term can be done using a process very much like long division of whole numbers. You must be careful to subtract entire expressions, not just the first term. Stop when the degree of the remainder is less than the degree of the divisor. The remainder can be written using R notation or as a fraction added to the quotient with the remainder in the numerator and the divisor in the denominator.<\/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-2590\">\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\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Divide a Degree 3 Polynomial by a Degree 1 Polynomial (Long Division with Missing Term). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Rxds7Q_UTeo\">https:\/\/youtu.be\/Rxds7Q_UTeo<\/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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. <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><li>Ex 1: Divide a Trinomial by a Binomial Using Long Division. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/KUPFg__Djzw\">https:\/\/youtu.be\/KUPFg__Djzw<\/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>Ex 6: Divide a Polynomial by a Degree Two Binomial Using Long Division. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/P6OTbUf8f60\">https:\/\/youtu.be\/P6OTbUf8f60<\/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":21,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Divide a Trinomial by a Binomial Using Long Division\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/KUPFg__Djzw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Divide a Degree 3 Polynomial by a Degree 1 Polynomial (Long Division with Missing Term)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Rxds7Q_UTeo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 6: Divide a Polynomial by a Degree Two Binomial Using Long Division\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/P6OTbUf8f60\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"fb2fd6ff-a26d-48d9-9ee4-bcbaeea28ee3","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2590","chapter","type-chapter","status-publish","hentry"],"part":994,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2590","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2590\/revisions"}],"predecessor-version":[{"id":5585,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2590\/revisions\/5585"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/parts\/994"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapters\/2590\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/media?parent=2590"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2590"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/contributor?post=2590"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/wp-json\/wp\/v2\/license?post=2590"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}