{"id":125584,"date":"2023-03-22T01:52:00","date_gmt":"2023-03-21T20:22:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125584"},"modified":"2023-11-22T12:58:11","modified_gmt":"2023-11-22T07:28:11","slug":"rd-sharma-class-9-solutions-chapter-6-exercise-6-5","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-5\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125588\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-Exercise-6.5.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-Exercise-6.5.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-Exercise-6.5-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5:<\/strong> Prepare for your upcoming Maths tests and exa<strong>ms <\/strong>with 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>. You will see how easy and self-explanatory solutions are designed by our subject matter experts. You can 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 6.5 from the link given in this 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-69d80ff141e47\" 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-69d80ff141e47\"><\/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-6-5\/#download-rd-sharma-class-9-solutions-chapter-6-exercise-65-pdf\" title=\"Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5 PDF\">Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5 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-6-5\/#access-answers-of-rd-sharma-class-9-solutions-chapter-6-exercise-65\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5\">Access answers of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5<\/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-6-5\/#faqs-on-rd-sharma-class-9-solutions-chapter-6-exercise-65\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5\">FAQs on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5<\/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-6-5\/#from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-for-chapter-6-exercise-65\" title=\"From where can I download the PDF of RD Sharma Class 9 Solutions for Chapter 6 Exercise 6.5?\">From where can I download the PDF of RD Sharma Class 9 Solutions for Chapter 6 Exercise 6.5?<\/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-6-5\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-for-chapter-6-exercise-65\" title=\"How much does it cost to download the PDF of RD Sharma Class 9 Solutions for Chapter 6 Exercise 6.5?\">How much does it cost to download the PDF of RD Sharma Class 9 Solutions for Chapter 6 Exercise 6.5?<\/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-6-5\/#can-i-access-the-rd-sharma-class-9-solutions-chapter-6-exercise-65-pdf-offline\" title=\"Can I access the RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5\u00a0PDF offline?\">Can I access the RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-6-exercise-65-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5 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-6.5.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions for Chapter 6 Exercise 6.5<\/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-6.5.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-6-exercise-65\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Using the factor theorem, factorize each of the following polynomials:<\/p>\n<p><strong>Question 1: x<sup>3<\/sup>&nbsp;+ 6x<sup>2<\/sup>&nbsp;+ 11x + 6<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Let f(x) = x<sup>3<\/sup>&nbsp;+ 6x<sup>2<\/sup>&nbsp;+ 11x + 6<\/p>\n<p>Step 1: Find the factors of the constant term<\/p>\n<p>Here constant term = 6<\/p>\n<p>Factors of 6 are \u00b11, \u00b12, \u00b13, \u00b16<\/p>\n<p>Step 2: Find the factors of f(x)<\/p>\n<p>Let x + 1 = 0<\/p>\n<p>\u21d2 x = -1<\/p>\n<p>Put the value of x in f(x)<\/p>\n<p>f(-1) = (\u22121)<sup>3<\/sup>&nbsp;+ 6(\u22121)<sup>2<\/sup>&nbsp;+ 11(\u22121) + 6<\/p>\n<p>= -1 + 6 -11 + 6<\/p>\n<p>= 12 \u2013 12<\/p>\n<p>= 0<\/p>\n<p>So, (x + 1) is the factor of f(x)<\/p>\n<p>Let x + 2 = 0<\/p>\n<p>\u21d2 x = -2<\/p>\n<p>Put the value of x in f(x)<\/p>\n<p>f(-2) = (\u22122)<sup>3<\/sup>&nbsp;+ 6(\u22122)<sup>2<\/sup>&nbsp;+ 11(\u22122) + 6 = -8 + 24 \u2013 22 + 6 = 0<\/p>\n<p>So, (x + 2) is the factor of f(x)<\/p>\n<p>Let x + 3 = 0<\/p>\n<p>\u21d2 x = -3<\/p>\n<p>Put the value of x in f(x)<\/p>\n<p>f(-3) = (\u22123)<sup>3<\/sup>&nbsp;+ 6(\u22123)<sup>2<\/sup>&nbsp;+ 11(\u22123) + 6 = -27 + 54 \u2013 33 + 6 = 0<\/p>\n<p>So, (x + 3) is the factor of f(x)<\/p>\n<p>Hence, f(x) = (x + 1)(x + 2)(x + 3)<\/p>\n<p><strong>Question 2: x<sup>3<\/sup>&nbsp;+ 2x<sup>2<\/sup>&nbsp;\u2013 x \u2013 2<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Let f(x) = x<sup>3<\/sup>&nbsp;+ 2x<sup>2<\/sup>&nbsp;\u2013 x \u2013 2<\/p>\n<p>Constant term = -2<\/p>\n<p>Factors of -2 are \u00b11, \u00b12<\/p>\n<p>Let x \u2013 1 = 0<\/p>\n<p>\u21d2 x = 1<\/p>\n<p>Put the value of x in f(x)<\/p>\n<p>f(1) = (1)<sup>3<\/sup>&nbsp;+ 2(1)<sup>2<\/sup>&nbsp;\u2013 1 \u2013 2 = 1 + 2 \u2013 1 \u2013 2 = 0<\/p>\n<p>So, (x \u2013 1) is factor of f(x)<\/p>\n<p>Let x + 1 = 0<\/p>\n<p>\u21d2 x = -1<\/p>\n<p>Put the value of x in f(x)<\/p>\n<p>f(-1) = (-1)<sup>3<\/sup>&nbsp;+ 2(-1)<sup>2<\/sup>&nbsp;\u2013 1 \u2013 2 = -1 + 2 + 1 \u2013 2 = 0<\/p>\n<p>(x + 1) is a factor of f(x)<\/p>\n<p>Let x + 2 = 0<\/p>\n<p>\u21d2 x = -2<\/p>\n<p>Put the value of x in f(x)<\/p>\n<p>f(-2) = (-2)<sup>3<\/sup>&nbsp;+ 2(-2)<sup>2<\/sup>&nbsp;\u2013 (-2) \u2013 2 = -8 + 8 + 2 \u2013 2 = 0<\/p>\n<p>(x + 2) is a factor of f(x)<\/p>\n<p>Let x \u2013 2 = 0<\/p>\n<p>\u21d2 x = 2<\/p>\n<p>Put the value of x in f(x)<\/p>\n<p>f(2) = (2)<sup>3<\/sup>&nbsp;+ 2(2)<sup>2<\/sup>&nbsp;\u2013 2 \u2013 2 = 8 + 8 \u2013 2 \u2013 2 = 12 \u2260 0<\/p>\n<p>(x \u2013 2) is not a factor of f(x)<\/p>\n<p>Hence f(x) = (x + 1)(x- 1)(x+2)<\/p>\n<p><strong>Question 3: x<sup>3<\/sup>&nbsp;\u2013 6x<sup>2<\/sup>&nbsp;+ 3x + 10<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Let f(x) = x<sup>3<\/sup>&nbsp;\u2013 6x<sup>2<\/sup>&nbsp;+ 3x + 10<\/p>\n<p>Constant term = 10<\/p>\n<p>Factors of 10 are \u00b11, \u00b12, \u00b15, \u00b110<\/p>\n<p>Let x + 1 = 0 or x = -1<\/p>\n<p>f(-1) = (-1)<sup>3<\/sup>&nbsp;\u2013 6(-1)<sup>2<\/sup>&nbsp;+ 3(-1) + 10 = 10 \u2013 10 = 0<\/p>\n<p>f(-1) = 0<\/p>\n<p>Let x + 2 = 0 or x = -2<\/p>\n<p>f(-2) = (-2)<sup>3<\/sup>&nbsp;\u2013 6(-2)<sup>2<\/sup>&nbsp;+ 3(-2) + 10 = -8 \u2013 24 \u2013 6 + 10 = -28<\/p>\n<p>f(-2) \u2260 0<\/p>\n<p>Let x \u2013 2 = 0 or x = 2<\/p>\n<p>f(2) = (2)<sup>3<\/sup>&nbsp;\u2013 6(2)<sup>2<\/sup>&nbsp;+ 3(2) + 10 = 8 \u2013 24 + 6 + 10 = 0<\/p>\n<p>f(2) = 0<\/p>\n<p>Let x \u2013 5 = 0 or x = 5<\/p>\n<p>f(5) = (5)<sup>3<\/sup>&nbsp;\u2013 6(5)<sup>2<\/sup>&nbsp;+ 3(5) + 10 = 125 \u2013 150 + 15 + 10 = 0<\/p>\n<p>f(5) = 0<\/p>\n<p>Therefore, (x + 1), (x \u2013 2) and (x-5) are factors of f(x)<\/p>\n<p>Hence f(x) = (x + 1) (x \u2013 2) (x-5)<\/p>\n<p><strong>Question 4: x<sup>4<\/sup>&nbsp;\u2013 7x<sup>3<\/sup>&nbsp;+ 9x<sup>2<\/sup>&nbsp;+ 7x- 10<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let f(x) = x<sup>4<\/sup>&nbsp;\u2013 7x<sup>3<\/sup>&nbsp;+ 9x<sup>2<\/sup>&nbsp;+ 7x- 10<\/p>\n<p>Constant term = -10<\/p>\n<p>Factors of -10 are \u00b11, \u00b12, \u00b15, \u00b110<\/p>\n<p>Let x \u2013 1 = 0 or x = 1<\/p>\n<p>f(1) = (1)<sup>4<\/sup>&nbsp;\u2013 7(1)<sup>3<\/sup>&nbsp;+ 9(1)<sup>2<\/sup>&nbsp;+ 7(1) \u2013 10 = 1 \u2013 7 + 9 + 7 -10 = 0<\/p>\n<p>f(1) = 0<\/p>\n<p>Let x + 1 = 0 or x = -1<\/p>\n<p>f(-1) = (-1)<sup>4<\/sup>&nbsp;\u2013 7(-1)<sup>3<\/sup>&nbsp;+ 9(-1)<sup>2<\/sup>&nbsp;+ 7(-1) \u2013 10 = 1 + 7 + 9 \u2013 7 -10 = 0<\/p>\n<p>f(-1) = 0<\/p>\n<p>Let x \u2013 2 = 0 or x = 2<\/p>\n<p>f(2) = (2)<sup>4<\/sup>&nbsp;\u2013 7(2)<sup>3<\/sup>&nbsp;+ 9(2)<sup>2<\/sup>&nbsp;+ 7(2) \u2013 10 = 16 \u2013 56 + 36 + 14 \u2013 10 = 0<\/p>\n<p>f(2) = 0<\/p>\n<p>Let x \u2013 5 = 0 or x = 5<\/p>\n<p>f(5) = (5)<sup>4<\/sup>&nbsp;\u2013 7(5)<sup>3<\/sup>&nbsp;+ 9(5)<sup>2<\/sup>&nbsp;+ 7(5) \u2013 10 = 625 \u2013 875 + 225 + 35 \u2013 10 = 0<\/p>\n<p>f(5) = 0<\/p>\n<p>Therefore, (x \u2013 1), (x + 1), (x \u2013 2) and (x-5) are factors of f(x)<\/p>\n<p>Hence f(x) = (x \u2013 1) (x + 1) (x \u2013 2) (x-5)<\/p>\n<p><strong>Question 5: x<sup>4<\/sup>&nbsp;\u2013 2x<sup>3<\/sup>&nbsp;\u2013 7x<sup>2<\/sup>&nbsp;+ 8x + 12<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>f(x) = x<sup>4<\/sup>&nbsp;\u2013 2x<sup>3<\/sup>&nbsp;\u2013 7x<sup>2<\/sup>&nbsp;+ 8x + 12<\/p>\n<p>Constant term = 12<\/p>\n<p>Factors of 12 are \u00b11, \u00b12, \u00b13, \u00b14, \u00b16, \u00b112<\/p>\n<p>Let x \u2013 1 = 0 or x = 1<\/p>\n<p>f(1) = (1)<sup>4<\/sup>&nbsp;\u2013 2(1)<sup>3<\/sup>&nbsp;\u2013 7(1)<sup>2<\/sup>&nbsp;+ 8(1) + 12 = 1 \u2013 2 \u2013 7 + 8 + 12 = 12<\/p>\n<p>f(1) \u2260 0<\/p>\n<p>Let x + 1 = 0 or x = -1<\/p>\n<p>f(-1) = (-1)<sup>4<\/sup>&nbsp;\u2013 2(-1)<sup>3<\/sup>&nbsp;\u2013 7(-1)<sup>2<\/sup>&nbsp;+ 8(-1) + 12 = 1 + 2 \u2013 7 \u2013 8 + 12 = 0<\/p>\n<p>f(-1) = 0<\/p>\n<p>Let x +2 = 0 or x = -2<\/p>\n<p>f(-2) = (-2)<sup>4<\/sup>&nbsp;\u2013 2(-2)<sup>3<\/sup>&nbsp;\u2013 7(-2)<sup>2<\/sup>&nbsp;+ 8(-2) + 12 = 16 + 16 \u2013 28 \u2013 16 + 12 = 0<\/p>\n<p>f(-2) = 0<\/p>\n<p>Let x \u2013 2 = 0 or x = 2<\/p>\n<p>f(2) = (2)<sup>4<\/sup>&nbsp;\u2013 2(2)<sup>3<\/sup>&nbsp;\u2013 7(2)<sup>2<\/sup>&nbsp;+ 8(2) + 12 = 16 \u2013 16 \u2013 28 + 16 + 12 = 0<\/p>\n<p>f(2) = 0<\/p>\n<p>Let x \u2013 3 = 0 or x = 3<\/p>\n<p>f(3) = (3)<sup>4<\/sup>&nbsp;\u2013 2(3)<sup>3<\/sup>&nbsp;\u2013 7(3)<sup>2<\/sup>&nbsp;+ 8(3) + 12 = 0<\/p>\n<p>f(3) = 0<\/p>\n<p>Therefore, (x + 1), (x + 2), (x \u2013 2) and (x-3) are factors of f(x)<\/p>\n<p>Hence f(x) = (x + 1)(x + 2) (x \u2013 2) (x-3)<\/p>\n<p><strong>Question 6: x<sup>4<\/sup>&nbsp;+ 10x<sup>3<\/sup>&nbsp;+ 35x<sup>2<\/sup>&nbsp;+ 50x + 24<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Let f(x) = x<sup>4<\/sup>&nbsp;+ 10x<sup>3<\/sup>&nbsp;+ 35x<sup>2<\/sup>&nbsp;+ 50x + 24<\/p>\n<p>Constant term = 24<\/p>\n<p>Factors of 24 are \u00b11, \u00b12, \u00b13, \u00b14, \u00b16, \u00b18, \u00b112, \u00b124<\/p>\n<p>Let x + 1 = 0 or x = -1<\/p>\n<p>f(-1) = (-1)<sup>4<\/sup>&nbsp;+ 10(-1)<sup>3<\/sup>&nbsp;+ 35(-1)<sup>2<\/sup>&nbsp;+ 50(-1) + 24 = 1 \u2013 10 + 35 \u2013 50 + 24 = 0<\/p>\n<p>f(1) = 0<\/p>\n<p>(x + 1) is a factor of f(x)<\/p>\n<p>Likewise, (x + 2),(x + 3),(x + 4) are also the factors of f(x)<\/p>\n<p>Hence f(x) = (x + 1) (x + 2)(x + 3)(x + 4)<\/p>\n<p><strong>Question 7: 2x<sup>4<\/sup>&nbsp;\u2013 7x<sup>3<\/sup>&nbsp;\u2013 13x<sup>2<\/sup>&nbsp;+ 63x \u2013 45<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let f(x) = 2x<sup>4<\/sup>&nbsp;\u2013 7x<sup>3<\/sup>&nbsp;\u2013 13x<sup>2<\/sup>&nbsp;+ 63x \u2013 45<\/p>\n<p>Constant term = -45<\/p>\n<p>Factors of -45 are \u00b11, \u00b13, \u00b15, \u00b19, \u00b115, \u00b145<\/p>\n<p>Here coefficient of x^4 is 2. So possible rational roots of f(x) are<\/p>\n<p>\u00b11, \u00b13, \u00b15, \u00b19, \u00b115, \u00b145, \u00b11\/2,\u00b13\/2,\u00b15\/2,\u00b19\/2,\u00b115\/2,\u00b145\/2<\/p>\n<p>Let x \u2013 1 = 0 or x = 1<\/p>\n<p>f(1) = 2(1)<sup>4<\/sup>&nbsp;\u2013 7(1)<sup>3<\/sup>&nbsp;\u2013 13(1)<sup>2<\/sup>&nbsp;+ 63(1) \u2013 45 = 2 \u2013 7 \u2013 13 + 63 \u2013 45 = 0<\/p>\n<p>f(1) = 0<\/p>\n<p>f(x) can be written as,<\/p>\n<p>f(x) = (x-1) (2x<sup>3<\/sup>&nbsp;\u2013 5x<sup>2<\/sup>&nbsp;-18x +45)<\/p>\n<p>or f(x) =(x-1)g(x) \u2026(1)<\/p>\n<p>Let x \u2013 3 = 0 or x = 3<\/p>\n<p>f(3) = 2(3)<sup>4<\/sup>&nbsp;\u2013 7(3)<sup>3<\/sup>&nbsp;\u2013 13(3)<sup>2<\/sup>&nbsp;+ 63(3) \u2013 45 = = 162 \u2013 189 \u2013 117 + 189 \u2013 45= 0<\/p>\n<p>f(3) = 0<\/p>\n<p>Now, we are available with 2 factors of f(x), (x \u2013 1) and (x \u2013 3)<\/p>\n<p>Here g(x) = 2x<sup>2<\/sup>&nbsp;(x-3) + x(x-3) -15(x-3)<\/p>\n<p>Taking (x-3) as common<\/p>\n<p>= (x-3)(2x<sup>2<\/sup>&nbsp;+ x \u2013 15)<\/p>\n<p>=(x-3)(2x<sup>2<\/sup>+6x \u2013 5x -15)<\/p>\n<p>= (x-3)(2x-5)(x+3)<\/p>\n<p>= (x-3)(x+3)(2x-5) \u2026.(2)<\/p>\n<p>From (1) and (2)<\/p>\n<p>f(x) =(x-1) (x-3)(x+3)(2x-5)<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-solutions-chapter-6-exercise-65\"><\/span>FAQs on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5<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-1631104553147\" 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-for-chapter-6-exercise-65\"><\/span>From where can I download the PDF of RD Sharma Class 9 Solutions for Chapter 6 Exercise 6.5?<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 id=\"faq-question-1631104584284\" 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-for-chapter-6-exercise-65\"><\/span>How much does it cost to download the PDF of RD Sharma Class 9 Solutions for Chapter 6 Exercise 6.5?<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-1631104618153\" 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-65-pdf-offline\"><\/span>Can I access the RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5\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>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5: Prepare for your upcoming Maths tests and exams with the RD Sharma Solutions Class 9 Maths. You will see how easy and self-explanatory solutions are designed by our subject matter experts. You can download the Free PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-5\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.5 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":125588,"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\/125584"}],"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=125584"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125584\/revisions"}],"predecessor-version":[{"id":510789,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125584\/revisions\/510789"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125588"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125584"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125584"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125584"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}