{"id":126007,"date":"2023-03-27T01:57:00","date_gmt":"2023-03-26T20:27:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126007"},"modified":"2026-05-02T06:29:43","modified_gmt":"2026-05-02T00:59:43","slug":"rd-sharma-class-9-solutions-chapter-6-mcqs","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-mcqs\/","title":{"rendered":"RD Sharma Class 9 Solutions: Complete Guide [2026]"},"content":{"rendered":"<p><script type=\"application\/ld+json\">\n{\n  \"@context\": \"https:\/\/schema.org\",\n  \"@type\": \"Article\",\n  \"headline\": \"RD Sharma Class 9 Solutions: Complete Guide [2026]\",\n  \"description\": \"Comprehensive guide to RD Sharma Class 9 Solutions for Maths, including chapter-wise solutions and MCQs for the 2026 academic year.\",\n  \"author\": {\n    \"@type\": \"Organization\",\n    \"name\": \"KopyKitab\"\n  },\n  \"publisher\": {\n    \"@type\": \"Organization\",\n    \"name\": \"KopyKitab\",\n    \"url\": \"https:\/\/www.kopykitab.com\"\n  },\n  \"datePublished\": \"2026-05-02\",\n  \"dateModified\": \"2026-05-02\"\n}\n<\/script><\/p>\n<div class=\"freshness-block\" style=\"background:#e8f5e9;padding:15px;border-left:4px solid #4caf50;margin:20px 0;border-radius:4px;\">\n<strong>Last Updated:<\/strong> May 02, 2026 | This article has been updated with the latest information for 2026.\n<\/div>\n<p><img alt=\"RD Sharma Class 9 Solutions Chapter 6 MCQs\" class=\"alignnone size-full wp-image-126063\" height=\"675\" loading=\"eager\" sizes=\"(max-width: 1200px) 100vw, 1200px\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-MCQS.jpg\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-MCQS.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-6-MCQS-768x432.jpg 768w\" width=\"1200\"\/><\/p>\n<p><strong>Read more:<\/strong> <a href=\"https:\/\/www.kopykitab.com\/blog\/category\/rd-sharma-solutions\/\">RD Sharma \u2014 Complete Guide<\/a><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 6 MCQs: <\/strong>Class 9 mein maths ka topper banna hai? Perfect study material choose karna bohot zaroori hai, aur iske liye aap hamesha <a href=\"\/blog\/rd-sharma-class-9-solutions-for-maths\/\" rel=\"noopener\" target=\"_blank\">RD Sharma Solutions Class 9 Maths<\/a> par bharosa kar sakte hain. Chahe aapke class tests hon ya assignments, humara detailed solutions guide hamesha aapke saath hai. RD Sharma Class 9 Solutions Chapter 6 MCQs ke baare mein complete information ke liye, poora blog padhiye.<\/p>\n<div class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\" id=\"ez-toc-container\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<p><span class=\"ez-toc-title-toggle\"><a aria-label=\"ez-toc-toggle-icon-1\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" href=\"#\"><label aria-label=\"Table of Content\" for=\"item-69f54baa7f295\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg class=\"list-377408\" fill=\"none\" height=\"20px\" style=\"fill: #000000;color:#000000\" viewbox=\"0 0 24 24\" width=\"20px\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg baseprofile=\"tiny\" class=\"arrow-unsorted-368013\" height=\"10px\" style=\"fill: #000000;color:#000000\" version=\"1.2\" viewbox=\"0 0 24 24\" width=\"10px\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\"><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\"><\/path><\/svg><\/span><\/label><input id=\"item-69f54baa7f295\" type=\"checkbox\"\/><\/a><\/span><\/div>\n<nav>\n<ul class=\"ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default\">\n<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-mcqs\/#access-answers-of-rd-sharma-class-9-solutions-chapter-6-mcqs\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 6 MCQs\">Access answers of RD Sharma Class 9 Solutions Chapter 6 MCQs<\/a><\/li>\n<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-mcqs\/#faqs-on-rd-sharma-class-9-solutions-chapter-6-mcqs\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 6 MCqs\">FAQs on RD Sharma Class 9 Solutions Chapter 6 MCqs<\/a>\n<ul class=\"ez-toc-list-level-3\">\n<li class=\"ez-toc-heading-level-3\"><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-mcqs\/#how-many-questions-are-there-in-rd-sharma-class-9-solutions-chapter-6-mcqs\" title=\"How many questions are there in RD Sharma Class 9 Solutions Chapter 6 MCQs?\">How many questions are there in RD Sharma Class 9 Solutions Chapter 6 MCQs?<\/a><\/li>\n<li class=\"ez-toc-page-1 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-mcqs\/#is-it-even-beneficial-to-study-rd-sharma-class-9-solutions-chapter-6-mcqs\" title=\"Is it even beneficial to study RD Sharma Class 9 Solutions Chapter 6 MCQs?\">Is it even beneficial to study RD Sharma Class 9 Solutions Chapter 6 MCQs?<\/a><\/li>\n<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-mcqs\/#are-the-solutions-rd-sharma-class-9-solutions-chapter-6-mcqs-relevant\" title=\"Are the solutions RD Sharma Class 9 Solutions Chapter 6 MCQs relevant?\">Are the solutions RD Sharma Class 9 Solutions Chapter 6 MCQs relevant?<\/a><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<h2><span class=\"ez-toc-section\" id=\"complete-solutions-for-chapter-6-mcqs\"><\/span><span class=\"ez-toc-section\" id=\"access-answers-of-RD-sharma-class-9-solutions-chapter-6-mcqs\"><\/span><strong>Complete Solutions for Chapter 6 MCQs<\/strong><span class=\"ez-toc-section-end\"><\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>RD Sharma ki Class 9 Chapter 6 MCQs factorization of polynomials ke concept ko strong banane ke liye perfect hai. Yahan har question ka step-by-step solution diya gaya hai jo aapko concept clear karne mein help karega. Polynomials ka factorization ek fundamental topic hai jo aage ke chapters mein bhi kaam aayega.<\/p>\n<p>Is chapter mein total 20 MCQs hain jo different difficulty levels par based hain. Beginners se lekar advanced level tak ke students in questions se benefit le sakte hain. Har question ka detailed explanation diya gaya hai taaki aap method samajh sako.<\/p>\n<p><strong>Mark the correct alternative in each of the following:<br \/><\/strong><strong>Question 1.<br \/><\/strong><strong>If x \u2013 2 is a factor of x<sup>2<\/sup> + 3 ax \u2013 2a, then a =<\/strong><br \/><strong>(a) 2<\/strong><br \/><strong>(b) -2<\/strong><br \/><strong>(c) 1<\/strong><br \/><strong>(d) -1<\/strong><br \/><strong>Solution:<br \/><\/strong>\u2234 x \u2013 2 is a factor of<br \/>f(x) = x<sup>2<\/sup> + 3 ax \u2013 2a<br \/>\u2234 Remainder = 0<br \/>Let x \u2013 2 = 0, then x = 2<br \/>Now f(2) = (2)<sup>2<\/sup> + 3a x 2 \u2013 2a<br \/>= 4 + 6a \u2013 2a = 4 + 4a<br \/>\u2234 Remainder = 0<br \/>\u2234 4 + 4a = 0 \u21d2 4a = -4<br \/>\u21d2 a = <span class=\"MathJax\" id=\"MathJax-Element-1-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-1\"><span class=\"mrow\" id=\"MathJax-Span-2\"><span class=\"mfrac\" id=\"MathJax-Span-3\"><span class=\"mrow\" id=\"MathJax-Span-4\"><span class=\"mo\" id=\"MathJax-Span-5\">\u2212<\/span><span class=\"mn\" id=\"MathJax-Span-6\">4<\/span><\/span><span class=\"mn\" id=\"MathJax-Span-7\">4<\/span><\/span><\/span><\/span><\/span> = -1<br \/>\u2234 a= -1 <strong>(d)<\/strong><\/p>\n<p><strong>Question 2.<br \/><\/strong><strong>If x<sup>3<\/sup> + 6x<sup>2<\/sup> + 4x + k is exactly divisible by x + 2, then k =<\/strong><br \/><strong>(a) -6<\/strong><br \/><strong>(b) -7<\/strong><br \/><strong>(c) -8<\/strong><br \/><strong>(d) -10<br \/><\/strong><strong>Solution:<br \/><\/strong>f(x) \u2013 x<sup>3<\/sup> + 6x<sup>2<\/sup> + 4x + k is divisible by x + 2<br \/>\u2234 Remainder = 0<br \/>Let x + 2 = 0, then x = -2<br \/>\u2234 f(-2) = (-2)<sup>3<\/sup> + 6(-2)<sup>2<\/sup> + 4(-2) + k<br \/>= -8 + 24-8 + k = 8 + k<br \/>\u2234 x + 2 is a factor<br \/>\u2234 Remainder = 0.<br \/>\u21d2 8 + k= 0 \u21d2 k = -8<br \/>k = -8 <strong>(c)<\/strong><\/p>\n<div class=\"code-block code-block-2\"> <\/div>\n<p><strong>Question 3.<br \/><\/strong><strong>If x \u2013 a is a factor of x<sup>3<\/sup> \u2013 3x<sup>2<\/sup> a + 2a<sup>2<\/sup>x + b, then the value of b is<\/strong><br \/><strong>(a) 0<\/strong><br \/><strong>(b) 2<\/strong><br \/><strong>(c) 1<\/strong><br \/><strong>(d) 3<br \/><\/strong><strong>Solution:<br \/><\/strong>\u2234 x \u2013 a is a factor of x<sup>3<\/sup> \u2013 3x<sup>2<\/sup> a + 2a<sup>2<\/sup>x + b<br \/>Let f(x) = x<sup>3<\/sup> \u2013 3x<sup>2<\/sup> a + 2a<sup>2<\/sup>x+ b<br \/>and x \u2013 a = 0, then x = a<br \/>f(a) = a<sup>3<\/sup> \u2013 3a<sup>2<\/sup>.a + 2a<sup>2<\/sup>.a + b<br \/>= a<sup>3<\/sup> \u2013 3a<sup>3<\/sup> + 2a<sup>3<\/sup> + b = b<br \/>\u2235 x \u2013 a is a factor of f(x)<br \/>\u2234 b = 0 <strong>(a)<\/strong><\/p>\n<p><strong>Question 4.<br \/><\/strong><strong>If x<sup>140<\/sup> + 2x<sup>151<\/sup> + k is divisible by x + 1, then the value of k is<\/strong><br \/><strong>(a) 1<\/strong><br \/><strong>(b) -3<\/strong><br \/><strong>(c) 2<\/strong><br \/><strong>(d) -2<\/strong><br \/><strong>Solution:<br \/><\/strong>\u2234 x + 1 is a factor of f(x) = x<sup>140<\/sup> + 2x<sup>151<\/sup> + k<br \/>\u2234 The remainder will be zero<br \/>Let x + 1 = 0, then x = -1<br \/>\u2234 f(-1) = (-1)<sup>140<\/sup> + 2(-1)<sup>151<\/sup> + k<br \/>= 1 + 2 x (-1) + k {\u2235 140 is even and 151 is odd}<br \/>=1-2+k=k-1<br \/>\u2235 Remainder = 0<br \/>\u2234 k \u2013 1=0 \u21d2 k=1 <strong>(a)<\/strong><\/p>\n<p>Yahan ek important concept hai ki even powers mein negative numbers positive ho jaate hain, jabki odd powers mein negative hi rehte hain. Is concept ko yaad rakhna bohot zaroori hai MCQs solve karte waqt.<\/p>\n<p><strong>Question 5.<br \/><\/strong><strong>If x + 2 is a factor of x<sup>2<\/sup> + mx + 14, then m =<\/strong><br \/><strong>(a) 7<\/strong><br \/><strong>(b) 2<\/strong><br \/><strong>(c) 9<\/strong><br \/><strong>(d) 14<br \/><\/strong><strong>Solution:<br \/><\/strong>x + 2 is a factor of(x) = x<sup>2<\/sup> + mx + 14<br \/>Let x + 2 = 0, then x = -2<br \/>f(-2) = (-2)<sup>2<\/sup> + m{-2) + 14<br \/>= 4 \u2013 2m + 14 = 18 \u2013 2m<br \/>\u2234 x + 2 is a factor of f(x)<br \/>\u2234 Remainder = 0<br \/>\u21d2 18 \u2013 2m = 0<br \/>2m = 18 \u21d2 m = <span class=\"MathJax\" id=\"MathJax-Element-2-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-8\"><span class=\"mrow\" id=\"MathJax-Span-9\"><span class=\"mfrac\" id=\"MathJax-Span-10\"><span class=\"mn\" id=\"MathJax-Span-11\">18<\/span><span class=\"mn\" id=\"MathJax-Span-12\">2<\/span><\/span><\/span><\/span><\/span> = 9 <strong>(c)<\/strong><\/p>\n<p><strong>Question 6.<br \/>If x \u2013 3 is a factor of x<sup>2<\/sup> \u2013 ax \u2013 15, then a =<\/strong><br \/><strong>(a) -2<\/strong><br \/><strong>(b) 5<\/strong><br \/><strong>(c) -5<\/strong><br \/><strong>(d) 3<\/strong><br \/><strong>Solution:<br \/><\/strong>x \u2013 3 is a factor of(x) = x<sup>2<\/sup> \u2013 ax \u2013 15<br \/>Let x \u2013 3 = 0, then x = 3<br \/>\u2234 f(3) = (3)<sup>2<\/sup> \u2013 a(3) \u2013 15<br \/>= 9 -3a- 15<br \/>= -6 -3a<br \/>\u2234 x \u2013 3 is a factor<br \/>\u2234 Remainder = 0<br \/>-6 \u2013 3a = 0 \u21d2 3a = -6<br \/>\u2234 a = <span class=\"MathJax\" id=\"MathJax-Element-3-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-13\"><span class=\"mrow\" id=\"MathJax-Span-14\"><span class=\"mfrac\" id=\"MathJax-Span-15\"><span class=\"mrow\" id=\"MathJax-Span-16\"><span class=\"mo\" id=\"MathJax-Span-17\">\u2212<\/span><span class=\"mn\" id=\"MathJax-Span-18\">6<\/span><\/span><span class=\"mn\" id=\"MathJax-Span-19\">3<\/span><\/span><\/span><\/span><\/span> = -2 <strong>(a)<\/strong><\/p>\n<p><strong>Question 7.<br \/><\/strong><strong>If x<sup>51<\/sup> + 51 is divided by x + 1, the remainder is<\/strong><br \/><strong>(a) 0<\/strong><br \/><strong>(b) 1<\/strong><br \/><strong>(c) 49<\/strong><br \/><strong>(d) 50<br \/><\/strong><strong>Solution:<br \/><\/strong>Left(x) = x<sup>51<\/sup> + 51 is divisible by x + 1<br \/>Let x+1=0, then x = -1<br \/>\u2234 f(-1) = (-1)<sup>51<\/sup> + 51 =-1+51 (\u2235 power 51 is an odd integer)<br \/>= 50 <strong>(d)<\/strong><\/p>\n<p>Is question mein odd power ka concept use hua hai. Jab bhi negative number ka odd power hota hai, result negative aata hai. Yeh basic rule hai jo har student ko yaad rakhna chahiye.<\/p>\n<p><strong>Question 8.<br \/><\/strong><strong>If x+ 1 is a factor of the polynomial 2x<sup>2<\/sup> + kx, then k =<\/strong><br \/><strong>(a) -2<\/strong><br \/><strong>(b) -3<\/strong><br \/><strong>(c) 4<\/strong><br \/><strong>(d) 2<br \/><\/strong><strong>Solution:<br \/><\/strong>\u2234 x + 1 is a factor of the polynomial 2x<sup>2<\/sup> + kx<br \/>Let x+1=0, then x = -1<br \/>Now f(x) = 2x<sup>2<\/sup> + kx<br \/>\u2234 Remainder =f(-1) = 0<br \/>= 2(-1)<sup>2<\/sup> + k(-1)<br \/>= 2 x 1 + k x (-1) = 2 \u2013 k<br \/>\u2234 x + 1 is a factor of f(x)<br \/>\u2234 Remainder = 0<br \/>\u2234 2 \u2013 k = 0 \u21d2 k = 2 <strong>(d)<\/strong><\/p>\n<p><strong>Question 9.<br \/><\/strong><strong>If x + a is a factor of x<sup>4<\/sup> \u2013 a<sup>2<\/sup>x<sup>2<\/sup> + 3x \u2013 6a, then a =<\/strong><br \/><strong>(a) 0<\/strong><br \/><strong>(b) -1<\/strong><br \/><strong>(c) 1<\/strong><br \/><strong>(d) 2<br \/><\/strong><strong>Solution:<br \/><\/strong>x + a is a factor o f(x) = x<sup>4<\/sup>\u2013 a<sup>2<\/sup>x<sup>2<\/sup> + 3x \u2013 6a<br \/>Let x + a = 0, then x = -a<br \/>Now, f(-a) \u2013 (-a)<sup>4<\/sup> -a<sup>2<\/sup>(-a)<sup>2<\/sup> + 3 (-a) \u2013 6a<br \/>= a<sup>4<\/sup>-a<sup>4<\/sup>-3a-6a = -9a<br \/>\u2234 x + a is a factor of f(x)<br \/>\u2234Remainder = 0<br \/>\u2234 -9a = 0 \u21d2 a = 0 <strong>(a)<\/strong><\/p>\n<p>Yeh question thoda complex hai kyunki fourth power aur square terms dono involved hain. Dhyan se calculation karni padti hai aise questions mein.<\/p>\n<p><strong>Question 10.<br \/><\/strong><strong>The value of k for which x \u2013 1 is a factor of 4x<sup>3<\/sup> + 3x<sup>2<\/sup> \u2013 4x + k, is<\/strong><br \/><strong>(a) 3<\/strong><br \/><strong>(b) 1<\/strong><br \/><strong>(c) -2<\/strong><br \/><strong>(d) -3<br \/><\/strong><strong>Solution:<br \/><\/strong>x- 1 is a factor of f(x) = 4x<sup>3<\/sup> + 3x<sup>2<\/sup> \u2013 4x + k<br \/>Let x \u2013 1 = 0, then x = 1<br \/>f(1) = 4(1 )<sup>3<\/sup> + 3(1)<sup>2<\/sup> \u2013 4 x 1 + k<br \/>= 4+3-4+k=3+k<br \/>\u2234 x- 1 is a factor of f(x)<br \/>\u2234 Remainder = 0<br \/>\u2234 3 + k = 0 \u21d2 k = -3 <strong>(d)<\/strong><\/p>\n<p><strong>Question 11.<br \/><\/strong><strong>If x+2 and x-1 are the factors of x<sup>3<\/sup>+<\/strong> <strong>10x<sup>2<\/sup> + mx + n, then the values of m and n are respectively<\/strong><br \/><strong>(a) 5 and-3<\/strong><br \/><strong>(b) 17 and-8<\/strong><br \/><strong>(c) 7 and-18<\/strong><br \/><strong>(d) 23 and -19<\/strong><br \/><strong>Solution:<br \/><\/strong>x+ 2 and x \u2013 1 are the factors of<br \/>f(x) = x<sup>3<\/sup> + 10x<sup>2<\/sup> + mx + n<br \/>Let x + 2 = 0, then x = -2<br \/>\u2234 f(-2) = (-2)<sup>3<\/sup> + 10(-2)<sup>2<\/sup> + m(-2) + n<br \/>= -8 + 40 \u2013 2m + n = 32 \u2013 2m + n<br \/>\u2234 x + 2 is a factor of f(x)<br \/>\u2234 Remainder = 0<br \/>\u2234 32 \u2013 2m + n = 0 \u21d2 2m \u2013 n = 32 \u2026(i)<br \/>Again x \u2013 1 is a factor of f(x)<br \/>Let x-1=0, then x= 1<br \/>\u2234 f(1) = (1)<sup>3<\/sup> + 10(1)<sup>2<\/sup> + m x 1 +n<br \/>= 1 + 10+ m + n =m + n+11<br \/>\u2234 x- 1 is a factor of f(x)<br \/>\u2234 m + n+ 11=0 \u21d2 m+n =-11 \u2026(ii)<br \/>Adding (i) and (ii),<br \/>3m = 32 \u2013 11 = 21<br \/>\u21d2 m = <span class=\"MathJax\" id=\"MathJax-Element-4-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-20\"><span class=\"mrow\" id=\"MathJax-Span-21\"><span class=\"mfrac\" id=\"MathJax-Span-22\"><span class=\"mn\" id=\"MathJax-Span-23\">21<\/span><span class=\"mn\" id=\"MathJax-Span-24\">3<\/span><\/span><\/span><\/span><\/span> = 7<br \/>and n = -11 \u2013 m = -11 \u2013 7 = -18<br \/>\u2234 m= 7, n = -18 <strong>(c)<\/strong><\/p>\n<p>Is question mein do factors diye gaye hain, isliye do equations banaani padi. Simultaneous equations solve karna ek important skill hai mathematics mein.<\/p>\n<p><strong>Question 12.<br \/><\/strong>Let f(x) be a polynomial such that f( <span class=\"MathJax\" id=\"MathJax-Element-5-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-25\"><span class=\"mrow\" id=\"MathJax-Span-26\"><span class=\"mfrac\" id=\"MathJax-Span-27\"><span class=\"mrow\" id=\"MathJax-Span-28\"><span class=\"mo\" id=\"MathJax-Span-29\">\u2212<\/span><span class=\"mn\" id=\"MathJax-Span-30\">1<\/span><\/span><span class=\"mn\" id=\"MathJax-Span-31\">2<\/span><\/span><\/span><\/span><\/span> )= 0, then a factor of f(x) is<br \/>(a) 2x \u2013 1<br \/>(b) 2x + 1<br \/>(c) x- 1<br \/>(d) x + 1<br \/><strong>Solution:<\/strong><br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q12.1\" class=\"alignnone size-full wp-image-66100\" height=\"181\" loading=\"lazy\" sizes=\"(max-width: 357px) 100vw, 357px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q12.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q12.1.png 357w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q12.1-300x152.png 300w\" width=\"357\"\/><\/p>\n<p>Agar f(-1\/2) = 0 hai, toh iska matlab hai ki (x + 1\/2) ek factor hai, jo ki (2x + 1) ke equivalent hai. Yeh factor theorem ka direct application hai.<\/p>\n<p><strong>Question 13.<br \/><\/strong><strong>When x<sup>3<\/sup> \u2013 2x<sup>2<\/sup> + ax \u2013 b is divided by x<sup>2<\/sup> \u2013 2x-3, the remainder is x \u2013 6. The value of a and b are respectively.<\/strong><br \/><strong>(a) -2, -6<\/strong><br \/><strong>(b) 2 and -6<\/strong><br \/><strong>(c) -2 and 6<\/strong><br \/><strong>(d) 2 and 6<br \/><\/strong><strong>Solution:<br \/><\/strong>Let f(x) = x<sup>3<\/sup> \u2013 2x<sup>2<\/sup> + ax \u2013 b<br \/>and Dividing f(x) by x<sup>2<\/sup> \u2013 2x + 3<br \/>Remainder = x \u2013 6<br \/>Let p(x) = x<sup>3<\/sup> \u2013 2x<sup>2<\/sup> + ax \u2013 b \u2013 (x \u2013 6) or x<sup>3 <\/sup>\u2013 2x<sup>2<\/sup> + x(a \u2013 1) \u2013 b + 6 is divisible by x<sup>2<\/sup> \u2013 2x+ 3 exactly<br \/>Now, x<sup>2<\/sup>-2x-3 = x<sup>2<\/sup>-3x + x- 3<br \/>= x(x \u2013 3) + 1(x \u2013 3)<br \/>= (x \u2013 3) (x + 1)<br \/>\u2234 x \u2013 3 and x + 1 are the factors of p(x)<br \/>Let x \u2013 3 = 0, then x = 3<br \/>\u2234 p(3) = (3)<sup>3<\/sup> \u2013 2(3)<sup>2<\/sup> + (a-1)x3-b + 6<br \/>= 27-18 + 3a-3-b + 6<br \/>= 33-21+3 a-b<br \/>= 12 + 3 a-b<br \/>\u2234 x \u2013 3 is a factor<br \/>\u2234 12 + 3a-b = 0 \u21d2 3a-b = -12 \u2026(i)<br \/>Again let x + 1 = 0, then x = -1<br \/>\u2234p(-1) = (-1)<sup>3<\/sup> \u2013 2(-l)<sup>2<\/sup> + (a \u2013 1) X (-1) \u2013 b + 6<br \/>= -1-2 + 1 \u2013 a \u2013 6 + 6<br \/>= 4 \u2013 a- b<br \/>\u2234 x + 1 is a factor<br \/>4-a-b = 0 \u21d2 a + b = 4 \u2026(ii)<br \/>Adding (i) and (ii),<br \/>4 a = -12 + 4 = -8 \u21d2 a = <span class=\"MathJax\" id=\"MathJax-Element-6-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-32\"><span class=\"mrow\" id=\"MathJax-Span-33\"><span class=\"mfrac\" id=\"MathJax-Span-34\"><span class=\"mrow\" id=\"MathJax-Span-35\"><span class=\"mo\" id=\"MathJax-Span-36\">\u2212<\/span><span class=\"mn\" id=\"MathJax-Span-37\">8<\/span><\/span><span class=\"mn\" id=\"MathJax-Span-38\">4<\/span><\/span><\/span><\/span><\/span> = -2<br \/>From (ii),<br \/>and -2 + b = 4\u21d2 b = 4 + 2 = 6<br \/>\u2234 a = -2, b = 6 <strong>(c)<\/strong><\/p>\n<p>Yeh question polynomial division ke concept par based hai. Remainder theorem ka use karke hum unknown coefficients find kar sakte hain.<\/p>\n<p><strong>Question 14.<br \/><\/strong><strong>One factor of x<sup>4<\/sup> + x<sup>2<\/sup> \u2013 20 is x<sup>2<\/sup> + 5, the other is<\/strong><br \/><strong>(a) x<sup>2<\/sup> \u2013 4<\/strong><br \/><strong>(b) x \u2013 4<\/strong><br \/><strong>(c) x<sup>2<\/sup>-5<\/strong><br \/><strong>(d) x + 4<br \/><\/strong><strong>Solution:<br \/><\/strong><br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q14.1\" class=\"alignnone size-full wp-image-66101\" height=\"298\" loading=\"lazy\" sizes=\"(max-width: 365px) 100vw, 365px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q14.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q14.1.png 365w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q14.1-300x245.png 300w\" width=\"365\"\/><\/p>\n<p>Agar x<sup>4<\/sup> + x<sup>2<\/sup> \u2013 20 = (x<sup>2<\/sup> + 5)(x<sup>2<\/sup> \u2013 4) hai, toh dusra factor x<sup>2<\/sup> \u2013 4 hoga. Yeh simple division se pata chal jaata hai.<\/p>\n<p><strong>Question 15.<br \/><\/strong><strong>If (x \u2013 1) is a factor of polynomial fix) but not of g(x), then it must be a factor of<\/strong><br \/><strong>(a) f(x) g(x)<\/strong><br \/><strong>(b) -f(x) + g(x)<\/strong><br \/><strong>(c) f(x) \u2013 g(x)<\/strong><br \/><strong>(d) {f(x) + g(x)}g(x)<br \/><\/strong><strong>Solution:<br \/><\/strong>\u2234 (x \u2013 1) is a factor of a polynomial f(x)<br \/>But not of a polynomial g(x)<br \/>\u2234 (x \u2013 1) will be the factor of the product of f(x) and g(x) <strong>(a)<\/strong><\/p>\n<p>Factor properties ke according, agar ek factor kisi polynomial ka hai toh woh uske product mein bhi factor hoga.<\/p>\n<p><strong>Question 16.<br \/><\/strong><strong>(x + 1) is a factor of x<sup>n<\/sup> + 1 only if<\/strong><br \/><strong>(a) n is an odd integer<\/strong><br \/><strong>(b) n is an even integer<\/strong><br \/><strong>(c) n is a negative integer<\/strong><br \/><strong>(d) n is a positive integer<br \/><\/strong><strong>Solution:<br \/><\/strong>\u2234 (x + 1) is a factor of x<sup>n<\/sup>+ 1<br \/>Let x + 1 = 0, then x = -1<br \/>\u2234 f(x) = x<sup>n<\/sup> + 1<br \/>and f(-1) = (-1)<sup>n<\/sup> + 1<br \/>But (-1)<sup>n<\/sup> is positive if n is an even integer and negative if n is an odd integer and (-1)<sup>n<\/sup> +1=0 {\u2235 x + 1 is a factor of(x)}<br \/>(-1)<sup>n<\/sup> must be negative<br \/>\u2234 n is an odd integer <strong>(a)<\/strong><\/p>\n<p>Yahan even aur odd powers ka concept use hua hai. Odd powers mein negative sign preserve hota hai.<\/p>\n<p><strong>Question 17.<br \/><\/strong><strong>If x<sup>2<\/sup> + x + 1 is a factor of the polynomial 3x<sup>3<\/sup> + 8x<sup>2<\/sup> + 8x + 3 + 5k, then the value of k is<\/strong><br \/><strong>(a) 0<\/strong><br \/><strong>(b) <span class=\"MathJax\" id=\"MathJax-Element-7-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-39\"><span class=\"mrow\" id=\"MathJax-Span-40\"><span class=\"mfrac\" id=\"MathJax-Span-41\"><span class=\"mn\" id=\"MathJax-Span-42\">2<\/span><span class=\"mn\" id=\"MathJax-Span-43\">5<\/span><\/span><\/span><\/span><\/span><\/strong><br \/><strong>(c) <span class=\"MathJax\" id=\"MathJax-Element-8-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-44\"><span class=\"mrow\" id=\"MathJax-Span-45\"><span class=\"mfrac\" id=\"MathJax-Span-46\"><span class=\"mn\" id=\"MathJax-Span-47\">5<\/span><span class=\"mn\" id=\"MathJax-Span-48\">2<\/span><\/span><\/span><\/span><\/span><\/strong><br \/><strong>(d) -1<br \/><\/strong><strong>Solution:<br \/><\/strong>x<sup>2<\/sup> + x + 1 is a factor of<br \/>f(x) = 3x<sup>3<\/sup> + 8x<sup>2<\/sup> + 8x + 3 + 5k<br \/>Now dividing by x<sup>2<\/sup> + x + 1, we get<br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q17.1\" class=\"alignnone size-full wp-image-66102\" height=\"306\" loading=\"lazy\" sizes=\"(max-width: 361px) 100vw, 361px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q17.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q17.1.png 361w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q17.1-300x254.png 300w\" width=\"361\"\/><\/p>\n<p>Is question mein polynomial long division use karna pada hai kyunki quadratic factor diya gaya hai.<\/p>\n<p><strong>Question 18.<br \/><\/strong><strong>If (3x \u2013 1)<sup>7<\/sup> = a<sub>7<\/sub>x<sup><sub>7<\/sub><\/sup> + a<sub>6<\/sub>x<sup>6<\/sup> + a<sub>5<\/sub>x<sup>5<\/sup> + \u2026 + a<sub>1<\/sub>x + a<sub>0<\/sub>, then a<sub>7<\/sub> + a<sub>6<\/sub> + a<sub>5<\/sub> + \u2026 + a<sub>1<\/sub> + a<sub>0<\/sub> =<\/strong><br \/><strong>(a) 0<\/strong><br \/><strong>(b) 1<\/strong><br \/><strong>(c) 128<\/strong><br \/><strong>(d) 64<\/strong><br \/><strong>Solution:<br \/><\/strong>f(x) = [3(1) \u2013 1]<sup>7<\/sup> = a<sub>7<\/sub>x<sup>7<\/sup> + a<sub>6<\/sub>x<sup>6<\/sup> + a<sub>5<\/sub>x<sup>5<\/sup> + \u2026 + a<sub>1<\/sub>x + a<sub>0<br \/><\/sub>Let x = 1, then<br \/>f(1) = (3x- 1)<sup>7<\/sup> = a<sub>7<\/sub>(1)<sup>7<\/sup> + a<sub>6<\/sub>(1)<sup>6<\/sup> + a<sub>5<\/sub>(1)<sup>5<\/sup> + \u2026 + a<sub>1<\/sub> x 1 + a<sub>0<br \/><\/sub>\u21d2 (3 \u2013 1)<sup>7<\/sup> = a<sub>7<\/sub> x 1 + a<sub>6<\/sub> x 1+ a<sub>5<\/sub> x 1 + \u2026 + a<sub>1<\/sub>x 1 +a<sub>0<br \/><\/sub>\u21d2 (2)<sup>7<\/sup> = a<sub>7<\/sub> + a<sub>6<\/sub> + a<sub>5<\/sub> + \u2026 + a<sub>1<\/sub> +a<sub>0<\/sub><sup><br \/><\/sup>\u2234 a<sub>7<\/sub> + a<sub>6<\/sub> + a<sub>5<\/sub> + \u2026 + a<sub>1<\/sub> + a<sub>0<\/sub>= 128 <strong>(c)<\/strong><\/p>\n<p>Yeh binomial expansion ka interesting application hai. x = 1 substitute karke saare coefficients ka sum mil jaata hai.<\/p>\n<p><strong>Question 19.<br \/><\/strong><strong>If both x \u2013 2 and x \u2013 <span class=\"MathJax\" id=\"MathJax-Element-9-Frame\" tabindex=\"0\"><span class=\"math\" id=\"MathJax-Span-49\"><span class=\"mrow\" id=\"MathJax-Span-50\"><span class=\"mfrac\" id=\"MathJax-Span-51\"><span class=\"mn\" id=\"MathJax-Span-52\">1<\/span><span class=\"mn\" id=\"MathJax-Span-53\">2<\/span><\/span><\/span><\/span><\/span> are factors of px<sup>2<\/sup> + 5x + r, then<\/strong><br \/><strong>(a) p = r<\/strong><br \/><strong>(b) p + r = 0<\/strong><br \/><strong>(c) 2p + r = 0<\/strong><br \/><strong>(d) p + 2r = 0<br \/><\/strong><strong>Solution:<\/strong><br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q19.1\" class=\"alignnone size-full wp-image-66103\" height=\"516\" loading=\"lazy\" sizes=\"(max-width: 365px) 100vw, 365px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q19.1.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q19.1.png 365w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q19.1-212x300.png 212w\" width=\"365\"\/><br \/><img alt=\"RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q19.2\" class=\"alignnone size-full wp-image-66104\" height=\"166\" loading=\"lazy\" sizes=\"(max-width: 361px) 100vw, 361px\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q19.2.png\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q19.2.png 361w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-6-Factorisation-of-Polynomials-MCQS-Q19.2-300x138.png 300w\" width=\"361\"\/><\/p>\n<p>Do factors diye gaye hain, isliye do equations banegi aur unhe solve karke p aur r ke beech relation find karna hai.<\/p>\n<p><strong>Question 20.<br \/><\/strong><strong>If x<sup>2<\/sup> \u2013 1 is a factor of ax<sup>4<\/sup> + bx<sup>3<\/sup> + cx<sup>2<\/sup> + dx + e, then<\/strong><br \/><strong>(a) a + c + e- b + d<em><br \/>(b<\/em>)a + b + e = c + d<\/strong><br \/><strong><em>(c<\/em>)a + b + c = d+ e<em><br \/>(d<\/em>)b + c + d= a + e<\/strong><br \/><strong>Solution:<br \/><\/strong>X<sup>2<\/sup> \u2013 1 is a factor of ax<sup>4<\/sup> + bx<sup>3<\/sup> + cx<sup>2<\/sup> + dx + e<br \/>\u21d2 (x + 1), (x \u2013 1) are the factors of ax<sup>4<\/sup> + bx<sup>3 <\/sup>+ cx<sup>2<\/sup> + dx + e<br \/>Let f(x) = ax<sup>4<\/sup> + bx<sup>3<\/sup> + cx<sup>2<\/sup> + dx + e<br \/>and x + 1 = 0 then x = -1<br \/>\u2234 f(-1) = a(-1)<sup>4<\/sup> + b(-1)<sup>3<\/sup> + c(-1)<sup>2<\/sup> + d(-1) + e<br \/>= a- b + c- d+ e<br \/>\u2234 x + 1 is a factor of f(x)<br \/>\u2234 a-b + c- d+e = 0<br \/>\u21d2 a + c + e = b + d <strong>(a)<\/strong><\/p>\n<p>x<sup>2<\/sup> &#8211; 1 = (x+1)(x-1) hai, toh dono factors separately use karne padte hain. Yeh difference of squares ka concept hai.<\/p>\n<p>RD Sharma Class 9 Solutions Chapter 6 MCQs complete kar chuke hain. Ye saare questions <a href=\"https:\/\/www.cbse.gov.in\/\" rel=\"noopener\" target=\"_blank\">CBSE<\/a> Class 9 Maths exam ke liye bohot important hain. Factorization ke concepts ko strong banane ke liye regular practice zaroori hai.<\/p>\n<p>In MCQs mein jo main concepts cover hue hain woh hain:<\/p>\n<ul>\n<li><strong>Factor Theorem:<\/strong> Agar (x-a) polynomial f(x) ka factor hai toh f(a) = 0 hoga<\/li>\n<li><strong>Remainder Theorem:<\/strong> Polynomial ko (x-a) se divide karne par remainder f(a) hota hai<\/li>\n<li><strong>Even aur Odd Powers:<\/strong> Negative numbers ke even powers positive, odd powers negative hote hain<\/li>\n<li><strong>Polynomial Division:<\/strong> Long division method se quotient aur remainder find karna<\/li>\n<li><strong>Multiple Factors:<\/strong> Jab do ya zyada factors diye hon toh simultaneous equations solve karna<\/li>\n<\/ul>\n<p>Ye concepts aage ke chapters mein bhi use hote hain, especially quadratic equations aur coordinate geometry mein. Practice ke liye aur <a href=\"\/blog\/rd-sharma-solutions-class-9-maths-chapter-6-factorization-of-polynomials\/\" rel=\"noopener\" target=\"_blank\">RD Sharma Class 9 Solutions Chapter 6<\/a> ke exercises bhi solve kar sakte hain.<\/p>\n<div class=\"related-articles\" style=\"background:#fff3e0;padding:20px;border-left:4px solid #ff9800;margin:24px 0;border-radius:4px;\">\n<h3 style=\"margin:0 0 12px;color:#e65100;\"><span class=\"ez-toc-section\" id=\"you-may-also-like\"><\/span>You May Also Like<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul style=\"list-style:none;margin:0;padding:0;\">\n<li style=\"margin-bottom:8px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/category\/rd-sharma-solutions\/\" style=\"color:#1565c0;text-decoration:none;font-weight:500;\">\u2192 RD Sharma Solutions<\/a><\/li>\n<li style=\"margin-bottom:8px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/category\/rd-sharma-solutions\/\" style=\"color:#1565c0;text-decoration:none;font-weight:500;\">\u2192 Maths Solutions<\/a><\/li>\n<li style=\"margin-bottom:8px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/category\/cbse\/\" style=\"color:#1565c0;text-decoration:none;font-weight:500;\">\u2192 CBSE Preparation<\/a><\/li>\n<li style=\"margin-bottom:8px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/category\/exam-preparation\/\" style=\"color:#1565c0;text-decoration:none;font-weight:500;\">\u2192 Exam Preparation Guide<\/a><\/li>\n<\/ul>\n<\/div>\n<div class=\"cta-box\" style=\"background:linear-gradient(135deg,#e8f5e9,#c8e6c9);padding:24px;border-radius:8px;margin:24px 0;text-align:center;border:1px solid #a5d6a7;\">\n<h3 style=\"margin:0 0 8px;color:#2e7d32;\"><span class=\"ez-toc-section\" id=\"explore-study-materials\"><\/span>Explore Study Materials<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"margin:0 0 16px;color:#333;\">Get comprehensive study materials, practice tests, and expert guides for your exam preparation.<\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/competitive-exam-books\" style=\"display:inline-block;background:#2e7d32;color:#fff;padding:12px 32px;border-radius:6px;text-decoration:none;font-weight:600;\">Browse Study Materials \u2192<\/a>\n<\/div>\n<h2><span class=\"ez-toc-section\" id=\"frequently-asked-questions\"><\/span><span class=\"ez-toc-section\" id=\"faqs-on-RD-sharma-class-9-solutions-chapter-6-mcqs\"><\/span><strong>Frequently Asked Questions<\/strong><span class=\"ez-toc-section-end\"><\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div class=\"rank-math-block\" id=\"rank-math-faq\">\n<div class=\"rank-math-list\">\n<div class=\"rank-math-list-item\" id=\"faq-question-1631189499409\">\n<h3 class=\"rank-math-question\"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-class-9-solutions-chapter-6-mcqs\"><\/span><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-RD-sharma-class-9-solutions-chapter-6-mcqs\"><\/span>How many questions are there in RD Sharma Class 9 Solutions Chapter 6 MCQs?<span class=\"ez-toc-section-end\"><\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer\">\n<p>RD Sharma Class 9 Solutions Chapter 6 mein total 20 MCQs hain. Ye saare questions factorization of polynomials ke different concepts par based hain aur varying difficulty levels rakhte hain &#8211; basic se advanced tak.<\/p>\n<\/div>\n<\/div>\n<div class=\"rank-math-list-item\" id=\"faq-question-1631189545281\">\n<h3 class=\"rank-math-question\"><span class=\"ez-toc-section\" id=\"is-it-even-beneficial-to-study-rd-sharma-class-9-solutions-chapter-6-mcqs\"><\/span><span class=\"ez-toc-section\" id=\"is-it-even-beneficial-to-study-RD-sharma-class-9-solutions-chapter-6-mcqs\"><\/span>Is it even beneficial to study RD Sharma Class 9 Solutions Chapter 6 MCQs?<span class=\"ez-toc-section-end\"><\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer\">\n<p>Bilkul! RD Sharma ke MCQs aapki preparation ko bohot strong banate hain. Ye questions exam pattern ke according design kiye gaye hain aur concept clarity ke saath-saath problem-solving skills develop karte hain. Board exams mein aise hi questions aate hain.<\/p>\n<\/div>\n<\/div>\n<div class=\"rank-math-list-item\" id=\"faq-question-1631189572203\">\n<h3 class=\"rank-math-question\"><span class=\"ez-toc-section\" id=\"are-the-solutions-rd-sharma-class-9-solutions-chapter-6-mcqs-relevant\"><\/span><span class=\"ez-toc-section\" id=\"are-the-solutions-RD-sharma-class-9-solutions-chapter-6-mcqs-relevant\"><\/span>Are the solutions RD Sharma Class 9 Solutions Chapter 6 MCQs relevant?<span class=\"ez-toc-section-end\"><\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer\">\n<p>Haan, ye solutions bilkul relevant hain kyunki subject matter experts ne inhe design kiya hai. 2026 ke liye updated solutions provide kiye gaye hain jo current CBSE syllabus ke according hain. Step-by-step explanations se concept samajhna easy ho jaata hai.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Last Updated: May 02, 2026 | This article has been updated with the latest information for 2026. Read more: RD Sharma \u2014 Complete Guide RD Sharma Class 9 Solutions Chapter 6 MCQs: Class 9 mein maths ka topper banna hai? Perfect study material choose karna bohot zaroori hai, aur iske liye aap hamesha RD Sharma &#8230; <a title=\"RD Sharma Class 9 Solutions: Complete Guide [2026]\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-6-mcqs\/\" aria-label=\"More on RD Sharma Class 9 Solutions: Complete Guide [2026]\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":126063,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[61598,115469],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126007"}],"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=126007"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126007\/revisions"}],"predecessor-version":[{"id":575270,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126007\/revisions\/575270"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126063"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126007"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126007"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126007"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}