{"id":125529,"date":"2023-09-12T01:07:00","date_gmt":"2023-09-11T19:37:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125529"},"modified":"2023-09-13T21:09:21","modified_gmt":"2023-09-13T15:39:21","slug":"rd-sharma-class-9-solutions-chapter-6-exercise-6-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 (Updated for 2023)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125531\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-Exercise-6.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-Exercise-6.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-Exercise-6.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2: <\/strong>Download the Free PDF of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-6-factorization-of-polynomials\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 6<\/a> Exercise 6.2 from the link given in this blog. &nbsp;In this exercise, students will learn how to find the value of a polynomial. This study material helps students to master a concept by solving various questions listed in the exercise. Be it your Maths exam or class assignment, we have got you covered. All the solutions in the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> are designed by subject matter experts and are very reliable. To know more, read the whole blog.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69dc638318fdf\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69dc638318fdf\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-2\/#download-rd-sharma-class-9-solutions-chapter-6-exercise-62-pdf\" title=\"Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 PDF\">Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-2\/#access-answers-of-rd-sharma-class-9-solutions-chapter-6-exercise-62\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2\">Access answers of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-2\/#faqs-on-rd-sharma-class-9-solutions-chapter-6-exercise-62\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2\">FAQs on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-2\/#can-i-access-the-rd-sharma-class-9-solutions-chapter-6-exercise-62-pdf-offline\" title=\"Can I access the RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2\u00a0PDF offline?\">Can I access the RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2\u00a0PDF offline?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-2\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-6-exercise-62\" title=\"How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2?\">How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-2\/#from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-6-exercise-62\" title=\"From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2?\">From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-6-exercise-62-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-6-Ex-6.2.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-6-Ex-6.2.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-6-exercise-62\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: If f(x) = 2x<sup>3<\/sup>&nbsp;\u2013 13x<sup>2<\/sup>&nbsp;+ 17x + 12, find<\/strong><\/p>\n<p><strong>(i) f (2)<\/strong><\/p>\n<p><strong>(ii) f (-3)<\/strong><\/p>\n<p><strong>(iii) f(0)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>f(x) = 2x<sup>3<\/sup>&nbsp;\u2013 13x<sup>2<\/sup>&nbsp;+ 17x + 12<\/p>\n<p><strong>(i)<\/strong>&nbsp;f(2) = 2(2)<sup>3<\/sup>&nbsp;\u2013 13(2)<sup>&nbsp;2<\/sup>&nbsp;+ 17(2) + 12<\/p>\n<p>= 2 x 8 \u2013 13 x 4 + 17 x 2 + 12<\/p>\n<p>= 16 \u2013 52 + 34 + 12<\/p>\n<p>= 62 \u2013 52<\/p>\n<p>= 10<\/p>\n<p><strong>(ii)<\/strong>&nbsp;f(-3) = 2(-3)<sup>3<\/sup>&nbsp;\u2013 13(-3)<sup>&nbsp;2<\/sup>&nbsp;+ 17 x (-3) + 12<\/p>\n<p>= 2 x (-27) \u2013 13 x 9 + 17 x (-3) + 12<\/p>\n<p>= -54 \u2013 117 -51 + 12<\/p>\n<p>= -222 + 12<\/p>\n<p>= -210<\/p>\n<p><strong>(iii)<\/strong>&nbsp;f(0) = 2 x (0)<sup>3<\/sup>&nbsp;\u2013 13(0)<sup>&nbsp;2<\/sup>&nbsp;+ 17 x 0 + 12<\/p>\n<p>= 0-0 + 0+ 12<\/p>\n<p>= 12<\/p>\n<p><strong>Question 2: Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:<\/strong><\/p>\n<p><strong>(i) f(x) = 3x + 1, x = \u22121\/3<\/strong><\/p>\n<p><strong>(ii) f(x) = x<sup>2<\/sup>&nbsp;\u2013 1, x = 1,\u22121<\/strong><\/p>\n<p><strong>(iii) g(x) = 3x<sup>2<\/sup>&nbsp;\u2013 2 , x = 2\/\u221a3 , \u22122\/\u221a3<\/strong><\/p>\n<p><strong>(iv) p(x) = x<sup>3<\/sup>&nbsp;\u2013 6x<sup>2<\/sup>&nbsp;+ 11x \u2013 6 , x = 1, 2, 3<\/strong><\/p>\n<p><strong>(v) f(x) = 5x \u2013 \u03c0, x = 4\/5<\/strong><\/p>\n<p><strong>(vi) f(x) = x<sup>2<\/sup>&nbsp;, x = 0<\/strong><\/p>\n<p><strong>(vii) f(x) = lx + m, x = \u2212m\/l<\/strong><\/p>\n<p><strong>(viii) f(x) = 2x + 1, x = 1\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>&nbsp;f(x) = 3x + 1, x = \u22121\/3<\/p>\n<p>f(x) = 3x + 1<\/p>\n<p>Substitute x = \u22121\/3 in f(x)<\/p>\n<p>f( \u22121\/3) = 3(\u22121\/3) + 1<\/p>\n<p>= -1 + 1<\/p>\n<p>= 0<\/p>\n<p>Since, the result is 0, so x = \u22121\/3 is the root of 3x + 1<\/p>\n<p><strong>(ii)<\/strong>&nbsp;f(x) = x<sup>2<\/sup>&nbsp;\u2013 1, x = 1,\u22121<\/p>\n<p>f(x) = x<sup>2<\/sup>&nbsp;\u2013 1<\/p>\n<p>Given that x = (1 , -1)<\/p>\n<p>Substitute x = 1 in f(x)<\/p>\n<p>f(1) = 1<sup>2<\/sup>&nbsp;\u2013 1<\/p>\n<p>= 1 \u2013 1<\/p>\n<p>= 0<\/p>\n<p>Now, substitute x = (-1) in f(x)<\/p>\n<p>f(-1) = (\u22121)<sup>2<\/sup>&nbsp;\u2013 1<\/p>\n<p>= 1 \u2013 1<\/p>\n<p>= 0<\/p>\n<p>Since , the results when x = 1 and x = -1 are 0, so (1 , -1) are the roots of the polynomial f(x) = x<sup>2&nbsp;<\/sup>\u2013 1<\/p>\n<p><strong>(iii)<\/strong>&nbsp;g(x) = 3x<sup>2<\/sup>&nbsp;\u2013 2 , x = 2\/\u221a3 , \u22122\/\u221a3<\/p>\n<p>g(x) = 3x<sup>2<\/sup>&nbsp;\u2013 2<\/p>\n<p>Substitute x = 2\/\u221a3 in g(x)<\/p>\n<p>g(2\/\u221a3) = 3(2\/\u221a3)<sup>2<\/sup>&nbsp;\u2013 2<\/p>\n<p>= 3(4\/3) \u2013 2<\/p>\n<p>= 4 \u2013 2<\/p>\n<p>= 2 \u2260 0<\/p>\n<p>Now, Substitute x = \u22122\/\u221a3 in g(x)<\/p>\n<p>g(2\/\u221a3) = 3(-2\/\u221a3)<sup>2<\/sup>&nbsp;\u2013 2<\/p>\n<p>= 3(4\/3) \u2013 2<\/p>\n<p>= 4 \u2013 2<\/p>\n<p>= 2 \u2260 0<\/p>\n<p>Since, the results when x = 2\/\u221a3 and x = \u22122\/\u221a3) are not 0. Therefore (2\/\u221a3 , \u22122\/\u221a3 ) are not zeros of 3x<sup>2<\/sup>\u20132.<\/p>\n<p><strong>(iv)<\/strong>&nbsp;p(x) = x<sup>3<\/sup>&nbsp;\u2013 6x<sup>2<\/sup>&nbsp;+ 11x \u2013 6 , x = 1, 2, 3<\/p>\n<p>p(1) = 1<sup>3<\/sup>&nbsp;\u2013 6(1)<sup>2<\/sup>&nbsp;+ 11x 1 \u2013 6 = 1 \u2013 6 + 11 \u2013 6 = 0<\/p>\n<p>p(2) = 2<sup>3<\/sup>&nbsp;\u2013 6(2)<sup>2<\/sup>&nbsp;+ 11\u00d72 \u2013 6 = 8 \u2013 24 + 22 \u2013 6 = 0<\/p>\n<p>p(3) = 3<sup>3<\/sup>&nbsp;\u2013 6(3)<sup>2<\/sup>&nbsp;+ 11\u00d73 \u2013 6 = 27 \u2013 54 + 33 \u2013 6 = 0<\/p>\n<p>Therefore, x = 1, 2, 3 are zeros of p(x).<\/p>\n<p><strong>(v)<\/strong>&nbsp;f(x) = 5x \u2013 \u03c0, x = 4\/5<\/p>\n<p>f(4\/5) = 5 x 4\/5 \u2013 \u03c0 = 4 \u2013 \u03c0 \u2260 0<\/p>\n<p>Therefore, x = 4\/5 is not a zero of f(x).<\/p>\n<p><strong>(vi)<\/strong>&nbsp;f(x) = x<sup>2<\/sup>&nbsp;, x = 0<\/p>\n<p>f(0) = 0<sup>2<\/sup>&nbsp;= 0<\/p>\n<p>Therefore, x = 0 is a zero of f(x).<\/p>\n<p><strong>(vii)<\/strong>&nbsp;f(x) = lx + m, x = \u2212m\/l<\/p>\n<p>f(\u2212m\/l) = l x \u2212m\/l + m = -m + m = 0<\/p>\n<p>Therefore, x = \u2212m\/l is a zero of f(x).<\/p>\n<p><strong>(viii)<\/strong>&nbsp;f(x) = 2x + 1, x = \u00bd<\/p>\n<p>f(1\/2) = 2x 1\/2 + 1 = 1 + 1 = 2 \u2260 0<\/p>\n<p>Therefore, x = \u00bd is not a zero of f(x).<\/p>\n<p>This is the complete blog on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths exam, ask in the comments.&nbsp;<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-solutions-chapter-6-exercise-62\"><\/span><strong>FAQs on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631101206185\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-class-9-solutions-chapter-6-exercise-62-pdf-offline\"><\/span>Can I access the RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631101261058\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-6-exercise-62\"><\/span>How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631101265292\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-6-exercise-62\"><\/span>From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2: Download the Free PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 from the link given in this blog. &nbsp;In this exercise, students will learn how to find the value of a polynomial. This study material helps students to master a concept by solving &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 (Updated for 2023)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.2 (Updated for 2023)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":125531,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[3086,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125529"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=125529"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125529\/revisions"}],"predecessor-version":[{"id":472236,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125529\/revisions\/472236"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125531"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125529"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125529"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125529"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}