{"id":4222,"date":"2023-02-02T06:00:55","date_gmt":"2023-02-02T06:00:55","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4222"},"modified":"2023-02-02T06:03:46","modified_gmt":"2023-02-02T06:03:46","slug":"ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-4\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.4"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 2 (Polynomials)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 2 Polynomials <\/strong>Exercise 2.4 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 2: Polynomials\u00a0<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-2-polynomials-ex-2-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 2.5<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 2.4\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Determine which of the following polynomials has (x + 1) a factor:<br \/>\n<\/strong><strong>(i) x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ x + 1<br \/>\n(ii) x<sup>4<\/sup>\u00a0+ x<sup>3<\/sup>\u00a0+ x<sup>2<\/sup>\u00a0+ x + 1<br \/>\n(iii) x<sup>4<\/sup>\u00a0+ 3x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup>\u00a0+ x + 1<br \/>\n(iv) x<sup>3<\/sup>\u00a0\u2013 x<sup>2<\/sup> \u2013 (2 + \u221a2 )x + \u221a2 <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<strong>(i) x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ x + 1<br \/>\n<\/strong>Let p(x) =\u00a0x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ x + 1<br \/>\nThe zero of x + 1 is -1.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [x + 1 = 0 means x = -1]<br \/>\np(\u22121) = (\u22121)<sup>3 <\/sup>+ (\u22121)<sup>2 <\/sup>+ (\u22121) + 1<br \/>\n= \u22121 + 1 \u2212 1 + 1<br \/>\n= 0<br \/>\n\u2234 By factor theorem, x + 1 is a factor of x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ x + 1.\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) x<sup>4 <\/sup>+ x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ x + 1<br \/>\n<\/strong>Let p(x) = x<sup>4 <\/sup>+ x<sup>3 <\/sup>+ x<sup>2 <\/sup>+ x + 1<br \/>\nThe zero of x + 1 is -1.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[x + 1 = 0 means x = -1]<br \/>\np(\u22121) = (\u22121)<sup>4 <\/sup>+ (\u22121)<sup>3 <\/sup>+ (\u22121)<sup>2 <\/sup>+ (\u22121) + 1<br \/>\n= 1 \u2212 1 + 1 \u2212 1 + 1<br \/>\n= 1 \u2260 0<br \/>\n\u2234 By factor theorem, x + 1 is not a factor of x<sup>4<\/sup>\u00a0+\u00a0x<sup>3<\/sup>\u00a0+\u00a0x<sup>2<\/sup> +\u00a0x\u00a0+ 1.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) x<sup>4 <\/sup>+ 3x<sup>3 <\/sup>+ 3x<sup>2 <\/sup>+ x + 1<br \/>\n<\/strong>Let p(x)=\u00a0x<sup>4 <\/sup>+ 3x<sup>3 <\/sup>+ 3x<sup>2 <\/sup>+ x + 1<br \/>\nThe zero of x + 1 is -1.<br \/>\np(\u22121) = (\u22121)<sup>4 <\/sup>+ 3(\u22121)<sup>3 <\/sup>+ 3(\u22121)<sup>2 <\/sup>+ (\u22121) + 1<br \/>\n=1 \u2212 3 + 3 \u2212 1 + 1<br \/>\n=1 \u2260 0<br \/>\n\u2234 By factor theorem, x+1 is not a factor of x<sup>4 <\/sup>+ 3x<sup>3 <\/sup>+ 3x<sup>2 <\/sup>+ x +1.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) x<sup>3\u00a0<\/sup>\u2013 x<sup>2<\/sup>\u2013 (2 + \u221a2)x + \u221a2<br \/>\n<\/strong>Let p(x) =\u00a0x<sup>3 <\/sup>\u2013 x<sup>2 <\/sup>\u2013 (2 + \u221a2)x + \u221a2<br \/>\nThe zero of x + 1 is -1.<br \/>\np(\u22121) = (-1)<sup>3 <\/sup>\u2013 (-1)<sup>2 <\/sup>\u2013 (2 + \u221a2)(-1) + \u221a2<br \/>\n= \u22121 \u2212 1 + 2 + \u221a2 + \u221a2<br \/>\n= 2\u221a2 \u2260 0<br \/>\n\u2234 By factor theorem, x + 1 is not a factor of x<sup>3 <\/sup>\u2013 x<sup>2 <\/sup>\u2013 (2 + \u221a2)x + \u221a2.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Use the Factor Theorem to determine whether\u00a0g(x) is a factor of\u00a0p(x) in each of the following cases:<br \/>\n<\/strong><strong>(i) p(x) = 2x<sup>3 <\/sup>+ x<sup>2 <\/sup>\u2013 2x \u2013 1, g(x) = x + 1<br \/>\n(ii) p(x)= x<sup>3<\/sup>\u00a0+ 3x<sup>2<\/sup>\u00a0+ 3x + 1, g (x) = x + 2<br \/>\n(iii) p (x) = x<sup>3<\/sup>\u00a0\u2013 4x<sup>2<\/sup>\u00a0+ x + 6, g (x) = x \u2013 3<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<strong>(i) p(x) = 2x<sup>3 <\/sup>+ x<sup>2 <\/sup>\u2013 2x \u2013 1, g(x) = x + 1<br \/>\n<\/strong>p(x) = 2x<sup>3 <\/sup>+ x<sup>2 <\/sup>\u2013 2x \u2013 1, g(x) = x+1<br \/>\ng(x) = 0<br \/>\n\u21d2 x + 1 = 0<br \/>\n\u21d2 x = \u22121<br \/>\n\u2234 Zero of g(x) is -1.<br \/>\nNow,<br \/>\np(\u22121) = 2(\u22121)<sup>3 <\/sup>+ (\u22121)<sup>2 <\/sup>\u2013 2(\u22121) \u2013 1<br \/>\n= \u22122 + 1 + 2 \u2212 1<br \/>\n= 0<br \/>\n\u2234 By factor theorem, g(x) is a factor of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) p(x) = x<sup>3 <\/sup>+ 3x<sup>2 <\/sup>+ 3x + 1, g(x) = x+2<br \/>\n<\/strong>g(x) = 0<br \/>\n\u21d2 x + 2 = 0<br \/>\n\u21d2 x = \u22122<br \/>\n\u2234 Zero of g(x) is -2.<br \/>\nNow,<br \/>\np(\u22122) = (\u22122)<sup>3 <\/sup>+ 3(\u22122)<sup>2 <\/sup>+ 3(\u22122) + 1<br \/>\n= \u22128 + 12 \u2212 6 + 1<br \/>\n= \u22121 \u2260 0<br \/>\n\u2234 By factor theorem, g(x) is not a factor of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) p(x) = x<sup>3 <\/sup>\u2013 4x<sup>2 <\/sup>+ x + 6, g(x) = x \u2013 3<br \/>\n<\/strong>g(x) = 0<br \/>\n\u21d2 x \u2212 3 = 0<br \/>\n\u21d2 x = 3<br \/>\n\u2234 Zero of g(x) is 3.<br \/>\nNow,<br \/>\np(3) = (3)<sup>3 <\/sup>\u2212 4(3)<sup>2 <\/sup>+ (3) + 6<br \/>\n= 27 \u2212 36 + 3 + 6<br \/>\n= 0<br \/>\n\u2234 By factor theorem, g(x) is a factor of p(x).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Find the value of\u00a0k, if\u00a0x\u20131 is a factor of\u00a0p(x) in each of the following cases:<br \/>\n<\/strong><strong>(i) p(x) = x<sup>2 <\/sup>+ x + k<br \/>\n(ii) p (x) = 2x<sup>2<\/sup>\u00a0+ kx + \u221a2<br \/>\n(iii) p (x) = kx<sup>2<\/sup>\u00a0\u2013 \u221a2 x + 1<br \/>\n(iv) p (x) = kx<sup>2<\/sup>\u00a0\u2013 3x + k<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<strong>(i) p(x) = x<sup>2 <\/sup>+ x + k<br \/>\n<\/strong>If x &#8211; 1 is a factor of p(x), then p(1) = 0<br \/>\nBy Factor Theorem<br \/>\n\u21d2 (1)<sup>2 <\/sup>+ (1) + k = 0<br \/>\n\u21d2 1 + 1 + k = 0<br \/>\n\u21d2 2 + k = 0<br \/>\n\u21d2 k = \u22122<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) p(x) = 2x<sup>2 <\/sup>+ kx + <\/strong>\u221a2<br \/>\nIf x &#8211; 1 is a factor of p(x), then p(1)=0<br \/>\n\u21d2 2(1)<sup>2 <\/sup>+ k(1) + \u221a2 = 0<br \/>\n\u21d2 2 + k + \u221a2 = 0<br \/>\n\u21d2 k = \u2212(2 + \u221a2)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) p(x) = kx<sup>2 <\/sup>\u2013 <\/strong>\u221a<strong>2x + 1<br \/>\n<\/strong>If x &#8211; 1 is a factor of p(x), then p(1)=0<br \/>\n\u21d2 k(1)<sup>2 <\/sup>&#8211; \u221a2(1) + 1 = 0<br \/>\n\u21d2 k = \u221a2 &#8211; 1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) p(x) = kx<sup>2 <\/sup>\u2013 3x + k<br \/>\n<\/strong>If x &#8211; 1 is a factor of p(x), then p(1) = 0<br \/>\n\u21d2 k(1)<sup>2 <\/sup>\u2013 3(1) + k = 0<br \/>\n\u21d2 k \u2212 3 + k = 0<br \/>\n\u21d2 2k \u2212 3 = 0<br \/>\n\u21d2 k= 3\/2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Factorize:<br \/>\n<\/strong><strong>(i) 12x<sup>2 <\/sup>\u2013 7x + 1<br \/>\n(ii) 2x<sup>2<\/sup>\u00a0+ 7x + 3<br \/>\n(iii) 6x<sup>2<\/sup>\u00a0+ 5x \u2013 6<br \/>\n(iv) 3x<sup>2<\/sup>\u00a0\u2013 x \u2013 4<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<\/strong><br \/>\n<strong>(i) 12x<sup>2 <\/sup>\u2013 7x + 1<br \/>\n<\/strong>\u21d2 12x<sup>2<\/sup> \u2013 4x &#8211; 3x + 1<br \/>\n\u21d2 4x (3x \u2013 1 ) -1 (3x \u2013 1)<br \/>\n\u21d2 (3x -1) (4x -1)<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 2x<sup>2<\/sup>+7x+3<br \/>\n<\/strong>\u21d2 2x<sup>2 <\/sup>+ 6x + 1x + 3<br \/>\n\u21d2 2x (x + 3) + 1(x + 3)<br \/>\n\u21d2 (2x + 1)(x + 3)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 6x<sup>2 <\/sup>+ 5x &#8211; 6<br \/>\n<\/strong>\u21d2 6x<sup>2 <\/sup>+ 9x \u2013 4x \u2013 6<br \/>\n\u21d2 3x(2x + 3) \u2013 2(2x + 3)<br \/>\n\u21d2 (2x + 3)(3x \u2013 2)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 3x<sup>2 <\/sup>\u2013 x \u2013 4<br \/>\n<\/strong>\u21d2 3x<sup>2 <\/sup>\u2013 4x + 3x \u2013 4<br \/>\n= x(3x \u2013 4) + 1(3x \u2013 4)<br \/>\n= (3x \u2013 4)(x + 1)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Factorize:<br \/>\n<\/strong><strong>(i) x<sup>3 <\/sup>\u2013 2x<sup>2 <\/sup>\u2013 x + 2<br \/>\n(ii) x<sup>3<\/sup>\u00a0\u2013 3x<sup>2<\/sup>\u00a0\u2013 9x \u2013 5<br \/>\n(iii) x<sup>3<\/sup>\u00a0+ 13x<sup>2<\/sup>\u00a0+ 32x + 20<br \/>\n(iv) 2y<sup>3<\/sup>\u00a0+ y<sup>2<\/sup>\u00a0\u2013 2y \u2013 1<br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;<br \/>\n(i) x<sup>3 <\/sup>\u2013 2x<sup>2 <\/sup>\u2013 x + 2<br \/>\n<\/strong>Rearranging the terms, we have x<sup>3<\/sup>\u00a0\u2013 x \u2013 2x<sup>2<\/sup>\u00a0+ 2<br \/>\n\u21d2 x(x<sup>2<\/sup>\u00a0\u2013 1) \u2013 2(x<sup>2<\/sup> -1)<br \/>\n\u21d2 (x<sup>2<\/sup>\u00a0\u2013 1)(x \u2013 2)<br \/>\n\u21d2 [(x)<sup>2<\/sup>\u00a0\u2013 (1)<sup>2<\/sup>](x \u2013 2)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [\u2235 (a<sup>2<\/sup>\u00a0\u2013 b<sup>2<\/sup>) = (a + b)(a-b)]<br \/>\n\u21d2 (x \u2013 1)(x + 1)(x \u2013 2)<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) x<sup>3 <\/sup>\u2013 3x<sup>2 <\/sup>\u2013 9x \u2013 5<br \/>\n<\/strong>\u21d2 x<sup>3<\/sup>\u00a0+ x<sup>2<\/sup>\u00a0\u2013 4x<sup>2<\/sup> \u2013 4x \u2013 5x \u2013 5<br \/>\n\u21d2 x<sup>2<\/sup>\u00a0(x + 1) \u2013 4x(x + 1) \u2013 5(x + 1)<br \/>\n\u21d2 (x + 1)(x<sup>2<\/sup>\u00a0\u2013 4x \u2013 5)<br \/>\n\u21d2 (x + 1)(x<sup>2<\/sup>\u00a0\u2013 5x + x \u2013 5)<br \/>\n\u21d2 (x + 1)[x(x \u2013 5) + 1(x \u2013 5)]<br \/>\n\u21d2 (x + 1)(x \u2013 5)(x + 1)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) x<sup>3 <\/sup>+ 13x<sup>2 <\/sup>+ 32x + 20<br \/>\n<\/strong>\u21d2 x<sup>3<\/sup>\u00a0+ x<sup>2<\/sup>\u00a0+ 12x<sup>2<\/sup>\u00a0+ 12x + 20x + 20<br \/>\n\u21d2 x<sup>2<\/sup>(x + 1) + 12x(x +1) + 20(x + 1)<br \/>\n\u21d2 (x + 1)(x<sup>2<\/sup>\u00a0+ 12x + 20)<br \/>\n\u21d2 (x + 1)(x<sup>2<\/sup>\u00a0+ 2x + 10x + 20)<br \/>\n\u21d2 (x + 1)[x(x + 2) + 10(x + 2)]<br \/>\n\u21d2 (x + 1)(x + 2)(x + 10)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 2y<sup>3 <\/sup>+ y<sup>2 <\/sup>\u2013 2y \u2013 1<br \/>\n<\/strong>\u21d2 2y<sup>3<\/sup>\u00a0\u2013 2y<sup>2<\/sup>\u00a0+ 3y<sup>2<\/sup>\u00a0\u2013 3y + y \u2013 1<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 2y<sup>2<\/sup>(y \u2013 1) + 3y(y \u2013 1) + 1(y \u2013 1)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (y \u2013 1)(2y<sup>2<\/sup>\u00a0+ 3y + 1)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (y \u2013 1)(2y<sup>2<\/sup>\u00a0+ 2y + y + 1)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (y \u2013 1)[2y(y + 1) + 1(y + 1)]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (y \u2013 1)(y + 1)(2y + 1)<\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 2 (Polynomials)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 2 Polynomials Exercise 2.4 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 2: Polynomials\u00a0 NCERT Solution Class 9 Maths Ex &#8211; 2.1 NCERT Solution Class 9 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[703,702,186,704,701,5],"class_list":["post-4222","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-2-polynomials-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-2-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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