{"id":681,"date":"2025-06-20T17:02:12","date_gmt":"2025-06-20T17:02:12","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=681"},"modified":"2025-08-28T12:17:52","modified_gmt":"2025-08-28T12:17:52","slug":"advanced-integration-techniques-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/advanced-integration-techniques-background-youll-need-1\/","title":{"raw":"Advanced Integration Techniques: Background You'll Need 1","rendered":"Advanced Integration Techniques: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Divide polynomial expressions using long division techniques<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Polynomial Long Division<\/h2>\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.\r\n\r\n<section class=\"textbox example\">For example, let\u2019s divide [latex]178[\/latex] by [latex]3[\/latex] using long division.<center>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img 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\" \/> Long division steps[\/caption]\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<\/center>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.\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 another example.\r\n\r\n<\/section>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.\r\n\r\n<section class=\"textbox example\">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:<center>\r\n\r\n[caption id=\"attachment_995\" align=\"aligncenter\" width=\"617\"]<img class=\"wp-image-995 size-full\" 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\" \/> Steps of a division problem[\/caption]\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<\/center>We 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&nbsp;\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img 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\" \/> Labeled aspects of an equation[\/caption]\r\n\r\n<\/section>Writing the result in this manner illustrates the <strong>division algorithm<\/strong>.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>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\r\n&nbsp;\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&nbsp;\r\n\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\n&nbsp;\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<\/section><section class=\"textbox questionHelp\"><strong>How To: Given a polynomial and a binomial, use long division to divide the polynomial by the binomial<\/strong>\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 binomial 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<\/section><section class=\"textbox example\">Divide [latex]6{x}^{3}+11{x}^{2}-31x+15[\/latex]\u00a0by [latex]3x - 2[\/latex].[reveal-answer q=\"850001\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"850001\"]<center>\r\n\r\n[caption id=\"attachment_997\" align=\"aligncenter\" width=\"716\"]<img class=\"wp-image-997 size-full\" 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\" \/> Steps of a division problem[\/caption]\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<\/center>There is a remainder of [latex]1[\/latex]. We can express the result as:\r\n<p style=\"text-align: center;\">[latex]\\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x - 2}=2{x}^{2}+5x - 7+\\frac{1}{3x - 2}[\/latex]<\/p>\r\n<strong>Analysis<\/strong>\r\n\r\nWe can check our work by using the Division Algorithm to rewrite the solution then multiplying.\r\n<p style=\"text-align: center;\">[latex]\\left(3x - 2\\right)\\left(2{x}^{2}+5x - 7\\right)+1=6{x}^{3}+11{x}^{2}-31x+15[\/latex]<\/p>\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 [latex]1[\/latex]<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]24621[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]13812[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li><span data-sheets-root=\"1\">Divide polynomial expressions using long division techniques<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Polynomial Long Division<\/h2>\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.<\/p>\n<section class=\"textbox example\">For example, let\u2019s divide [latex]178[\/latex] by [latex]3[\/latex] using long division.<\/p>\n<div style=\"text-align: center;\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption class=\"wp-caption-text\">Long division steps<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<\/div>\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 another example.<\/p>\n<\/section>\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.<\/p>\n<section class=\"textbox example\">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<div style=\"text-align: center;\">\n<figure id=\"attachment_995\" aria-describedby=\"caption-attachment-995\" style=\"width: 617px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-995 size-full\" 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\" \/><figcaption id=\"caption-attachment-995\" class=\"wp-caption-text\">Steps of a division problem<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<\/div>\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>&nbsp;<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption class=\"wp-caption-text\">Labeled aspects of an equation<\/figcaption><\/figure>\n<\/section>\n<p>Writing the result in this manner illustrates the <strong>division algorithm<\/strong>.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>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>&nbsp;<\/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>&nbsp;<\/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>&nbsp;<\/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<\/section>\n<section class=\"textbox questionHelp\"><strong>How To: Given a polynomial and a binomial, use long division to divide the polynomial by the binomial<\/strong><\/p>\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 binomial 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<\/section>\n<section class=\"textbox example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q850001\">Show Solution<\/button><\/p>\n<div id=\"q850001\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">\n<figure id=\"attachment_997\" aria-describedby=\"caption-attachment-997\" style=\"width: 716px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-997 size-full\" 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\" \/><figcaption id=\"caption-attachment-997\" class=\"wp-caption-text\">Steps of a division problem<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<p>There is a remainder of [latex]1[\/latex]. We can express the result as:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x - 2}=2{x}^{2}+5x - 7+\\frac{1}{3x - 2}[\/latex]<\/p>\n<p><strong>Analysis<\/strong><\/p>\n<p>We can check our work by using the Division Algorithm to rewrite the solution then multiplying.<\/p>\n<p style=\"text-align: center;\">[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 [latex]1[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm24621\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24621&theme=lumen&iframe_resize_id=ohm24621&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm13812\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=13812&theme=lumen&iframe_resize_id=ohm13812&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":541,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/681"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/681\/revisions"}],"predecessor-version":[{"id":2025,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/681\/revisions\/2025"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/541"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/681\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=681"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=681"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=681"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=681"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}