{"id":125566,"date":"2023-09-09T01:45:00","date_gmt":"2023-09-08T20:15:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125566"},"modified":"2023-11-28T10:17:00","modified_gmt":"2023-11-28T04:47:00","slug":"rd-sharma-class-9-solutions-chapter-6-exercise-6-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-4\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125580\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-Exercise-6.4.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-Exercise-6.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-Exercise-6.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4:\u00a0<\/strong>Ace your Class 9 Maths exam 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>. 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 6.4 can be downloaded from the link mentioned in our blog. 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-69d9f40070c30\" 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-69d9f40070c30\"><\/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-6-4\/#download-rd-sharma-class-9-solutions-chapter-6-exercise-64-pdf\" title=\"Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4 PDF\">Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4 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-6-4\/#access-answers-of-rd-sharma-class-9-solutions-chapter-6-exercise-64\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4\">Access answers of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4<\/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-6-4\/#faqs-on-rd-sharma-class-9-solutions-chapter-6-exercise-64\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4\">FAQs on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4<\/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-6-4\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-6-exercise-64\" title=\"How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4?\">How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4?<\/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-6-4\/#from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-6-exercise-64\" title=\"From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4?\">From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4?<\/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-6-4\/#can-i-access-the-rd-sharma-class-9-solutions-chapter-6-exercise-64-pdf-offline\" title=\"Can I access the RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4\u00a0PDF offline?\">Can I access the RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4\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-64-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4 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-6.4.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/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-6.4.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-6-exercise-64\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1-7)<\/strong><\/p>\n<p><strong>Question 1: f(x) = x<sup>3<\/sup>\u00a0\u2013 6x<sup>2<\/sup>\u00a0+ 11x \u2013 6; g(x) = x \u2013 3<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n<p>g(x) = x -3 = 0<\/p>\n<p>or x = 3<\/p>\n<p>Remainder = f(3)<\/p>\n<p>Now,<\/p>\n<p>f(3) = (3)<sup>3<\/sup>\u00a0\u2013 6(3)<sup>2<\/sup>\u00a0+11 x 3 \u2013 6<\/p>\n<p>= 27 \u2013 54 + 33 \u2013 6<\/p>\n<p>= 60 \u2013 60<\/p>\n<p>= 0<\/p>\n<p>Therefore, g(x) is a factor of f(x)<\/p>\n<p><strong>Question 2: f(x) = 3X<sup>4<\/sup>\u00a0+ 17x<sup>3<\/sup>\u00a0+ 9x<sup>2<\/sup>\u00a0\u2013 7x \u2013 10; g(x) = x + 5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n<p>g(x) = x + 5 = 0, then x = -5<\/p>\n<p>Remainder = f(-5)<\/p>\n<p>Now,<\/p>\n<p>f(3) = 3(-5)<sup>4<\/sup>\u00a0+ 17(-5)<sup>3<\/sup>\u00a0+ 9(-5)<sup>2<\/sup>\u00a0\u2013 7(-5) \u2013 10<\/p>\n<p>= 3 x 625 + 17 x (-125) + 9 x (25) \u2013 7 x (-5) \u2013 10<\/p>\n<p>= 1875 -2125 + 225 + 35 \u2013 10<\/p>\n<p>= 0<\/p>\n<p>Therefore, g(x) is a factor of f(x).<\/p>\n<p><strong>Question 3: f(x) = x<sup>5<\/sup>\u00a0+ 3x<sup>4<\/sup>\u00a0\u2013 x<sup>3<\/sup>\u00a0\u2013 3x<sup>2<\/sup>\u00a0+ 5x + 15, g(x) = x + 3<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n<p>g(x) = x + 3 = 0, then x = -3<\/p>\n<p>Remainder = f(-3)<\/p>\n<p>Now,<\/p>\n<p>f(-3) = (-3)<sup>5<\/sup>\u00a0+ 3(-3)<sup>4<\/sup>\u00a0\u2013 (-3)<sup>3<\/sup>\u00a0\u2013 3(-3)<sup>2<\/sup>\u00a0+ 5(-3) + 15<\/p>\n<p>= -243 + 3 x 81 -(-27)-3 x 9 + 5(-3) + 15<\/p>\n<p>= -243 +243 + 27-27- 15 + 15<\/p>\n<p>= 0<\/p>\n<p>Therefore, g(x) is a factor of f(x).<\/p>\n<p><strong>Question 4: f(x) = x<sup>3<\/sup>\u00a0\u2013 6x<sup>2<\/sup>\u00a0\u2013 19x + 84, g(x) = x \u2013 7<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n<p>g(x) = x \u2013 7 = 0, then x = 7<\/p>\n<p>Remainder = f(7)<\/p>\n<p>Now,<\/p>\n<p>f(7) = (7)<sup>3<\/sup>\u00a0\u2013 6(7)<sup>2<\/sup>\u00a0\u2013 19 x 7 + 84<\/p>\n<p>= 343 \u2013 294 \u2013 133 + 84<\/p>\n<p>= 343 + 84 \u2013 294 \u2013 133<\/p>\n<p>= 0<\/p>\n<p>Therefore, g(x) is a factor of f(x).<\/p>\n<p><strong>Question 5: f(x) = 3x<sup>3<\/sup>\u00a0+ x<sup>2<\/sup>\u00a0\u2013 20x + 12, g(x) = 3x \u2013 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n<p>g(x) = 3x \u2013 2 = 0, then x = 2\/3<\/p>\n<p>Remainder = f(2\/3)<\/p>\n<p>Now,<\/p>\n<p>f(2\/3) = 3(2\/3)\u00a0<sup>3<\/sup>\u00a0+ (2\/3)\u00a0<sup>2<\/sup>\u00a0\u2013 20(2\/3) + 12<\/p>\n<p>= 3 x 8\/27 + 4\/9 \u2013 40\/3 + 12<\/p>\n<p>= 8\/9 + 4\/9 \u2013 40\/3 + 12<\/p>\n<p>= 0\/9<\/p>\n<p>= 0<\/p>\n<p>Therefore, g(x) is a factor of f(x).<\/p>\n<p><strong>Question 6: f(x) = 2x<sup>3<\/sup>\u00a0\u2013 9x<sup>2<\/sup>\u00a0+ x + 12, g(x) = 3 \u2013 2x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n<p>g(x) = 3 \u2013 2x = 0, then x = 3\/2<\/p>\n<p>Remainder = f(3\/2)<\/p>\n<p>Now,<\/p>\n<p>f(3\/2) = 2(3\/2)<sup>3<\/sup>\u00a0\u2013 9(3\/2)<sup>2<\/sup>\u00a0+ (3\/2) + 12<\/p>\n<p>= 2 x 27\/8 \u2013 9 x 9\/4 + 3\/2 + 12<\/p>\n<p>= 27\/4 \u2013 81\/4 + 3\/2 + 12<\/p>\n<p>= 0\/4<\/p>\n<p>= 0<\/p>\n<p>Therefore, g(x) is a factor of f(x).<\/p>\n<p><strong>Question 7: f(x) = x<sup>3<\/sup>\u00a0\u2013 6x<sup>2<\/sup>\u00a0+ 11x \u2013 6, g(x) = x<sup>2<\/sup>\u00a0\u2013 3x + 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0.<\/p>\n<p>g(x) = 0<\/p>\n<p>or x<sup>2<\/sup>\u00a0\u2013 3x + 2 = 0<\/p>\n<p>x<sup>2<\/sup>\u00a0\u2013 x \u2013 2x + 2 = 0<\/p>\n<p>x(x \u2013 1) \u2013 2(x \u2013 1) = 0<\/p>\n<p>(x \u2013 1) (x \u2013 2) = 0<\/p>\n<p>Therefore x = 1 or x = 2<\/p>\n<p>Now,<\/p>\n<p>&#x200d;f(1) = (1)<sup>3<\/sup>\u00a0\u2013 6(1)<sup>2<\/sup>\u00a0+ 11(1) \u2013 6 = 1-6+11-6= 12- 12 = 0<\/p>\n<p>f(2) = (2)<sup>3<\/sup>\u00a0\u2013 6(2)<sup>2<\/sup>\u00a0+ 11(2) \u2013 6 = 8 \u2013 24 + 22 \u2013 6 = 30 \u2013 30 = 0<\/p>\n<p>=&gt; f(1) = 0 and f(2) = 0<\/p>\n<p>&#x200d;Which implies g(x) is factor of f(x).<\/p>\n<p><strong>Question 8: Show that (x \u2013 2), (x + 3) and (x \u2013 4) are factors of x<sup>3<\/sup>\u00a0\u2013 3x<sup>2<\/sup>\u00a0\u2013 10x + 24.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let f(x) = x<sup>3<\/sup>\u00a0\u2013 3x<sup>2<\/sup>\u00a0\u2013 10x + 24<\/p>\n<p>If x \u2013 2 = 0, then x = 2,<\/p>\n<p>If x + 3 = 0 then x = -3,<\/p>\n<p>and If x \u2013 4 = 0 then x = 4<\/p>\n<p>Now,<\/p>\n<p>f(2) = (2)<sup>3<\/sup>\u00a0\u2013 3(2)<sup>2<\/sup>\u00a0\u2013 10 x 2 + 24 = 8 \u2013 12 \u2013 20 + 24 = 32 \u2013 32 = 0<\/p>\n<p>&#x200d;f(-3) = (-3)<sup>3<\/sup>\u00a0\u2013 3(-3)<sup>2<\/sup>\u00a0\u2013 10 (-3) + 24 = -27 -27 + 30 + 24 = -54 + 54 = 0<\/p>\n<p>f(4) = (4)<sup>3<\/sup>\u00a0\u2013 3(4)<sup>2<\/sup>\u00a0\u2013 10 x 4 + 24 = 64-48 -40 + 24 = 88 \u2013 88 = 0<\/p>\n<p>f(2) = 0<\/p>\n<p>f(-3) = 0<\/p>\n<p>f(4) = 0<\/p>\n<p>Hence (x \u2013 2), (x + 3) and (x \u2013 4) are the factors of f(x)<\/p>\n<p><strong>Question 9: Show that (x + 4), (x \u2013 3) and (x \u2013 7) are factors of x<sup>3<\/sup>\u00a0\u2013 6x<sup>2<\/sup>\u00a0\u2013 19x + 84.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let f(x) = x<sup>3<\/sup>\u00a0\u2013 6x<sup>2<\/sup>\u00a0\u2013 19x + 84<\/p>\n<p>If x + 4 = 0, then x = -4<\/p>\n<p>If x \u2013 3 = 0, then x = 3<\/p>\n<p>and if x \u2013 7 = 0, then x = 7<\/p>\n<p>Now,<\/p>\n<p>f(-4) = (-4)<sup>3<\/sup>\u00a0\u2013 6(-4)<sup>2<\/sup>\u00a0\u2013 19(-4) + 84 = -64 \u2013 96 + 76 + 84 = 160 \u2013 160 = 0<\/p>\n<p>f(-4) = 0<\/p>\n<p>f(3) = (3)<sup>\u00a03<\/sup>\u00a0\u2013 6(3)<sup>\u00a02<\/sup>\u00a0\u2013 19 x 3 + 84 = 27 \u2013 54 \u2013 57 + 84 = 111 -111=0<\/p>\n<p>f(3) = 0<\/p>\n<p>f(7) = (7)<sup>\u00a03<\/sup>\u00a0\u2013 6(7)<sup>\u00a02<\/sup>\u00a0\u2013 19 x 7 + 84 = 343 \u2013 294 \u2013 133 + 84 = 427 \u2013 427 = 0<\/p>\n<p>f(7) = 0<\/p>\n<p>Hence (x + 4), (x \u2013 3), (x \u2013 7) are the factors of f(x).<\/p>\n<p>This is the complete blog on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-solutions-chapter-6-exercise-64\"><\/span><strong>FAQs on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4<\/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-1631103219404\" 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-64\"><\/span>How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4?<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-1631103259249\" 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-64\"><\/span>From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4?<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-1631103293765\" 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-64-pdf-offline\"><\/span>Can I access the RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4\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 6.4:\u00a0Ace your Class 9 Maths exam with the RD Sharma Solutions Class 9 Maths. The Free PDF of RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4 can be downloaded from the link mentioned in our blog. To know more, read the whole blog. Download RD Sharma &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-exercise-6-4\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 6 Exercise 6.4 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":125580,"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\/125566"}],"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=125566"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125566\/revisions"}],"predecessor-version":[{"id":513202,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125566\/revisions\/513202"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125580"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125566"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125566"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}