{"id":544444,"date":"2021-10-02T05:02:06","date_gmt":"2021-10-02T05:02:06","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=544444"},"modified":"2021-10-04T05:37:07","modified_gmt":"2021-10-04T05:37:07","slug":"rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/","title":{"rendered":"RD Sharma Solutions for Class 10 Maths Chapter 2\u2013Polynomials"},"content":{"rendered":"\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Class 10: Maths Chapter 2 solutions. Complete Class 10 Maths Chapter 2 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\">RD Sharma Solutions for Class 10 Maths Chapter 2\u2013Polynomials<\/h2>\n\n\n\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">RD Sharma 10th Maths Chapter 2, Class 10 Maths Chapter 2 solutions<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Exercise 2.1 Page No: 2.33<\/h3>\n\n\n\n<p><strong>1. Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:<\/strong><\/p>\n\n\n\n<p><strong>(i) f(x) = x<sup>2&nbsp;<\/sup>\u2013 2x \u2013 8<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>f(x) = x<sup>2&nbsp;<\/sup>\u2013 2x \u2013 8<\/p>\n\n\n\n<p>To find the zeros, we put f(x) = 0<\/p>\n\n\n\n<p>\u21d2 x<sup>2&nbsp;<\/sup>\u2013 2x \u2013 8 = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; x<sup>2<\/sup>&nbsp;\u2013 4x + 2x \u2013 8 = 0<\/p>\n\n\n\n<p>\u21d2 x(x \u2013 4) + 2(x \u2013 4) = 0<\/p>\n\n\n\n<p>\u21d2 (x \u2013 4)(x + 2) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>x = 4 and x = -2<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are 4 and -2.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of x \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>4 + (-2)= \u2013 (-2) \/ 1<\/p>\n\n\n\n<p>2 = 2<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>4 x (-2) = (-8) \/ 1<\/p>\n\n\n\n<p>-8 = -8<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(ii) g(s) = 4s<sup>2&nbsp;<\/sup>\u2013 4s + 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>g(s) = 4s<sup>2&nbsp;<\/sup>\u2013 4s + 1<\/p>\n\n\n\n<p>To find the zeros, we put g(s) = 0<\/p>\n\n\n\n<p>\u21d2 4s<sup>2&nbsp;<\/sup>\u2013 4s + 1 = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; 4s<sup>2<\/sup>&nbsp;\u2013 2s \u2013 2s + 1= 0<\/p>\n\n\n\n<p>\u21d2 &nbsp;2s(2s \u2013 1) \u2013 (2s \u2013 1)&nbsp;= 0<\/p>\n\n\n\n<p>\u21d2 (2s \u2013 1)(2s \u2013 1) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>s = 1\/2 and s = 1\/2<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are 1\/2 and 1\/2.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of s \/ coefficient of s<sup>2<\/sup><\/p>\n\n\n\n<p>1\/2 + 1\/2 = \u2013 (-4) \/ 4<\/p>\n\n\n\n<p>1 = 1<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of s<sup>2<\/sup><\/p>\n\n\n\n<p>1\/2 x 1\/2 = 1\/4<\/p>\n\n\n\n<p>1\/4 = 1\/4<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(iii) h(t)=t<sup>2&nbsp;<\/sup>\u2013 15<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>h(t) = t<sup>2&nbsp;<\/sup>\u2013 15 = t<sup>2&nbsp;<\/sup>+(0)t \u2013 15<\/p>\n\n\n\n<p>To find the zeros, we put h(t) = 0<\/p>\n\n\n\n<p>\u21d2 t<sup>2&nbsp;<\/sup>\u2013 15 = 0<\/p>\n\n\n\n<p>\u21d2 (t + \u221a15)(t \u2013 \u221a15)= 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>t = \u221a15 and t = -\u221a15<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are \u221a15 and -\u221a15.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of t \/ coefficient of t<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a15 + (-\u221a15) = \u2013 (0) \/ 1<\/p>\n\n\n\n<p>0 = 0<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of t<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a15 x (-\u221a15) = -15\/1<\/p>\n\n\n\n<p>-15 = -15<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(iv) f(x) = 6x<sup>2<\/sup>&nbsp;\u2013 3 \u2013 7x<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>f(x) = 6x<sup>2&nbsp;<\/sup>\u2013 3 \u2013 7x<\/p>\n\n\n\n<p>To find the zeros, we put f(x) = 0<\/p>\n\n\n\n<p>\u21d2 6x<sup>2&nbsp;<\/sup>\u2013 3 \u2013 7x = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; 6x<sup>2<\/sup>&nbsp;\u2013 9x + 2x \u2013 3 = 0<\/p>\n\n\n\n<p>\u21d2 3x(2x \u2013 3) + 1(2x \u2013 3) = 0<\/p>\n\n\n\n<p>\u21d2 (2x \u2013 3)(3x + 1) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>x = 3\/2 and x = -1\/3<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are 3\/2 and -1\/3.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of x \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>3\/2 + (-1\/3) = \u2013 (-7) \/ 6<\/p>\n\n\n\n<p>7\/6 = 7\/6<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>3\/2 x (-1\/3) = (-3) \/ 6<\/p>\n\n\n\n<p>-1\/2 = -1\/2<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(v) p(x) = x<sup>2&nbsp;<\/sup>+ 2\u221a2x \u2013 6<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>p(x) = x<sup>2&nbsp;<\/sup>+ 2\u221a2x \u2013 6<\/p>\n\n\n\n<p>To find the zeros, we put p(x) = 0<\/p>\n\n\n\n<p>\u21d2 x<sup>2&nbsp;<\/sup>+ 2\u221a2x \u2013 6 = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; x<sup>2<\/sup>&nbsp;+ 3\u221a2x \u2013 \u221a2x \u2013 6 = 0<\/p>\n\n\n\n<p>\u21d2 x(x + 3\u221a2) \u2013 \u221a2 (x + 3\u221a2) = 0<\/p>\n\n\n\n<p>\u21d2 (x \u2013 \u221a2)(x + 3\u221a2) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>x = \u221a2 and x = -3\u221a2<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are \u221a2 and -3\u221a2.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of x \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a2 + (-3\u221a2) = \u2013 (2\u221a2) \/ 1<\/p>\n\n\n\n<p>-2\u221a2 = -2\u221a2<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a2 x (-3\u221a2) = (-6) \/ 2\u221a2<\/p>\n\n\n\n<p>-3 x 2 = -6\/1<\/p>\n\n\n\n<p>-6 = -6<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(vi) q(x)=\u221a3x<sup>2&nbsp;<\/sup>+ 10x + 7\u221a3<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>q(x) = \u221a3x<sup>2&nbsp;<\/sup>+ 10x + 7\u221a3<\/p>\n\n\n\n<p>To find the zeros, we put q(x) = 0<\/p>\n\n\n\n<p>\u21d2 \u221a3x<sup>2&nbsp;<\/sup>+ 10x + 7\u221a3 = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; \u221a3x<sup>2<\/sup>&nbsp;+ 3x +7x + 7\u221a3x = 0<\/p>\n\n\n\n<p>\u21d2 \u221a3x(x + \u221a3) + 7 (x + \u221a3) = 0<\/p>\n\n\n\n<p>\u21d2 (x + \u221a3)(\u221a3x + 7) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>x = -\u221a3 and x = -7\/\u221a3<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are -\u221a3 and -7\/\u221a3.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of x \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>-\u221a3 + (-7\/\u221a3) = \u2013 (10) \/\u221a3<\/p>\n\n\n\n<p>(-3-7)\/ \u221a3 = -10\/\u221a3<\/p>\n\n\n\n<p>-10\/ \u221a3 = -10\/\u221a3<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>(-\u221a3) x (-7\/\u221a3) = (7\u221a3) \/ \u221a3<\/p>\n\n\n\n<p>7 = 7<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(vii) f(x) = x<sup>2&nbsp;<\/sup>\u2013 (\u221a3 + 1)x + \u221a3<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>f(x) = x<sup>2&nbsp;<\/sup>\u2013 (\u221a3 + 1)x + \u221a3<\/p>\n\n\n\n<p>To find the zeros, we put f(x) = 0<\/p>\n\n\n\n<p>\u21d2 x<sup>2&nbsp;<\/sup>\u2013 (\u221a3 + 1)x + \u221a3 = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; x<sup>2<\/sup>&nbsp;\u2013 \u221a3x \u2013 x + \u221a3 = 0<\/p>\n\n\n\n<p>\u21d2 x(x \u2013 \u221a3) \u2013 1 (x \u2013 \u221a3) = 0<\/p>\n\n\n\n<p>\u21d2 (x \u2013 \u221a3)(x \u2013 1) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>x = \u221a3 and x = 1<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are \u221a3 and 1.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of x \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a3 + 1 = \u2013 (-(\u221a3 +1)) \/ 1<\/p>\n\n\n\n<p>\u221a3 + 1 = \u221a3 +1<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>1 x \u221a3 = \u221a3 \/ 1<\/p>\n\n\n\n<p>\u221a3 = \u221a3<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(viii) g(x)=a(x<sup>2<\/sup>+1)\u2013x(a<sup>2<\/sup>+1)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>g(x) = a(x<sup>2<\/sup>+1)\u2013x(a<sup>2<\/sup>+1)<\/p>\n\n\n\n<p>To find the zeros, we put g(x) = 0<\/p>\n\n\n\n<p>\u21d2 a(x<sup>2<\/sup>+1)\u2013x(a<sup>2<\/sup>+1) = 0<\/p>\n\n\n\n<p>\u21d2 ax<sup>2<\/sup>&nbsp;+ a \u2212 a<sup>2<\/sup>x \u2013 x = 0<\/p>\n\n\n\n<p>\u21d2 ax<sup>2&nbsp;<\/sup>\u2212 a<sup>2<\/sup>x \u2013 x + a = 0<\/p>\n\n\n\n<p>\u21d2 ax(x \u2212 a) \u2212 1(x \u2013 a) = 0<\/p>\n\n\n\n<p>\u21d2 (x \u2013 a)(ax \u2013 1) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>x = a and x = 1\/a<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are a and 1\/a.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of x \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>a + 1\/a = \u2013 (-(a<sup>2&nbsp;<\/sup>+ 1)) \/ a<\/p>\n\n\n\n<p>(a<sup>2<\/sup>&nbsp;+ 1)\/a = (a<sup>2<\/sup>&nbsp;+ 1)\/a<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of x<sup>2<\/sup><\/p>\n\n\n\n<p>a x 1\/a = a \/ a<\/p>\n\n\n\n<p>1 = 1<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(ix) h(s) = 2s<sup>2&nbsp;<\/sup>\u2013 (1 + 2\u221a2)s + \u221a2<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>h(s) = 2s<sup>2&nbsp;<\/sup>\u2013 (1 + 2\u221a2)s + \u221a2<\/p>\n\n\n\n<p>To find the zeros, we put h(s) = 0<\/p>\n\n\n\n<p>\u21d2 2s<sup>2&nbsp;<\/sup>\u2013 (1 + 2\u221a2)s + \u221a2 = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; 2s<sup>2&nbsp;<\/sup>\u2013 2\u221a2s \u2013 s + \u221a2 = 0<\/p>\n\n\n\n<p>\u21d2 2s(s<sup>&nbsp;<\/sup>\u2013 \u221a2) -1(s \u2013 \u221a2) = 0<\/p>\n\n\n\n<p>\u21d2 (2s \u2013 1)(s \u2013 \u221a2) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>x = \u221a2 and x = 1\/2<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are \u221a3 and 1.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of s \/ coefficient of s<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a2 + 1\/2 = \u2013 (-(1 + 2\u221a2)) \/ 2<\/p>\n\n\n\n<p>(2\u221a2 + 1)\/2 = (2\u221a2 +1)\/2<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of s<sup>2<\/sup><\/p>\n\n\n\n<p>1\/2 x \u221a2 = \u221a2 \/ 2<\/p>\n\n\n\n<p>\u221a2 \/ 2 = \u221a2 \/ 2<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(x) f(v) = v<sup>2&nbsp;<\/sup>+ 4\u221a3v \u2013 15<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>f(v) = v<sup>2&nbsp;<\/sup>+ 4\u221a3v \u2013 15<\/p>\n\n\n\n<p>To find the zeros, we put f(v) = 0<\/p>\n\n\n\n<p>\u21d2 v<sup>2&nbsp;<\/sup>+ 4\u221a3v \u2013 15 = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; v<sup>2<\/sup>&nbsp;+ 5\u221a3v \u2013 \u221a3v \u2013 15 = 0<\/p>\n\n\n\n<p>\u21d2 v(v + 5\u221a3) \u2013 \u221a3 (v + 5\u221a3) = 0<\/p>\n\n\n\n<p>\u21d2 (v \u2013 \u221a3)(v + 5\u221a3) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>v = \u221a3 and v = -5\u221a3<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are \u221a3 and -5\u221a3.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of v \/ coefficient of v<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a3 + (-5\u221a3) = \u2013 (4\u221a3) \/ 1<\/p>\n\n\n\n<p>-4\u221a3 = -4\u221a3<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of v<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a3 x (-5\u221a3) = (-15) \/ 1<\/p>\n\n\n\n<p>-5 x 3 = -15<\/p>\n\n\n\n<p>-15 = -15<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(xi) p(y) = y<sup>2&nbsp;<\/sup>+ (3\u221a5\/2)y \u2013 5<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>p(y) = y<sup>2&nbsp;<\/sup>+ (3\u221a5\/2)y \u2013 5<\/p>\n\n\n\n<p>To find the zeros, we put f(v) = 0<\/p>\n\n\n\n<p>\u21d2 y<sup>2&nbsp;<\/sup>+ (3\u221a5\/2)y \u2013 5 = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; y<sup>2<\/sup>&nbsp;\u2013 \u221a5\/2 y + 2\u221a5y \u2013 5 = 0<\/p>\n\n\n\n<p>\u21d2 y(y \u2013 \u221a5\/2) + 2\u221a5 (y \u2013 \u221a5\/2) = 0<\/p>\n\n\n\n<p>\u21d2 (y + 2\u221a5)(y \u2013 \u221a5\/2) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>y = \u221a5\/2 and y = -2\u221a5<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are \u221a5\/2 and -2\u221a5.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of y \/ coefficient of y<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a5\/2 + (-2\u221a5) = \u2013 (3\u221a5\/2) \/ 1<\/p>\n\n\n\n<p>-3\u221a5\/2 = -3\u221a5\/2<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of y<sup>2<\/sup><\/p>\n\n\n\n<p>\u221a5\/2 x (-2\u221a5) = (-5) \/ 1<\/p>\n\n\n\n<p>\u2013 (\u221a5)<sup>2<\/sup>&nbsp;= -5<\/p>\n\n\n\n<p>-5 = -5<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>(xii) q(y) = 7y<sup>2&nbsp;<\/sup>\u2013 (11\/3)y \u2013 2\/3<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>q(y) = 7y<sup>2&nbsp;<\/sup>\u2013 (11\/3)y \u2013 2\/3<\/p>\n\n\n\n<p>To find the zeros, we put q(y) = 0<\/p>\n\n\n\n<p>\u21d2 7y<sup>2&nbsp;<\/sup>\u2013 (11\/3)y \u2013 2\/3 = 0<\/p>\n\n\n\n<p>\u21d2&nbsp; (21y<sup>2<\/sup>&nbsp;\u2013 11y -2)\/3 = 0<\/p>\n\n\n\n<p>\u21d2 21y<sup>2<\/sup>&nbsp;\u2013 11y \u2013 2 = 0<\/p>\n\n\n\n<p>\u21d2 21y<sup>2<\/sup>&nbsp;\u2013 14y + 3y \u2013 2 = 0<\/p>\n\n\n\n<p>\u21d2 7y(3y \u2013 2) \u2013 1(3y + 2) = 0<\/p>\n\n\n\n<p>\u21d2 (3y \u2013 2)(7y + 1) = 0<\/p>\n\n\n\n<p>This gives us 2 zeros, for<\/p>\n\n\n\n<p>y = 2\/3 and y = -1\/7<\/p>\n\n\n\n<p>Hence, the zeros of the quadratic equation are 2\/3 and -1\/7.<\/p>\n\n\n\n<p>Now, for verification<\/p>\n\n\n\n<p>Sum of zeros = \u2013 coefficient of y \/ coefficient of y<sup>2<\/sup><\/p>\n\n\n\n<p>2\/3 + (-1\/7) = \u2013 (-11\/3) \/ 7<\/p>\n\n\n\n<p>-11\/21 = -11\/21<\/p>\n\n\n\n<p>Product of roots = constant \/ coefficient of y<sup>2<\/sup><\/p>\n\n\n\n<p>2\/3 x (-1\/7) = (-2\/3) \/ 7<\/p>\n\n\n\n<p>\u2013 2\/21 = -2\/21<\/p>\n\n\n\n<p>Therefore, the relationship between zeros and their coefficients is verified.<\/p>\n\n\n\n<p><strong>2. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by factorization.<\/strong><\/p>\n\n\n\n<p><strong>(i) -8\/3 , 4\/3<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>A quadratic polynomial formed for the given sum and product of zeros is given by:<\/p>\n\n\n\n<p>f(x) = x<sup>2<\/sup>&nbsp;+ -(sum of zeros) x + (product of roots)<\/p>\n\n\n\n<p>Here, the sum of zeros is = -8\/3 and product of zero= 4\/3<\/p>\n\n\n\n<p>Thus,<\/p>\n\n\n\n<p>The required polynomial f(x) is,<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;\u2013 (-8\/3)x + (4\/3)<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;+ 8\/3x + (4\/3)<\/p>\n\n\n\n<p>So, to find the zeros we put f(x) = 0<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;+ 8\/3x + (4\/3) = 0<\/p>\n\n\n\n<p>\u21d2 3x<sup>2<\/sup>&nbsp;+ 8x + 4 = 0<\/p>\n\n\n\n<p>\u21d2 3x<sup>2<\/sup>&nbsp;+ 6x + 2x + 4 = 0<\/p>\n\n\n\n<p>\u21d2 3x(x + 2) + 2(x + 2) = 0<\/p>\n\n\n\n<p>\u21d2 (x + 2) (3x + 2) = 0<\/p>\n\n\n\n<p>\u21d2 (x + 2) = 0 and, or (3x + 2) = 0<\/p>\n\n\n\n<p>Therefore, the two zeros are -2 and -2\/3.<\/p>\n\n\n\n<p><strong>(ii) 21\/8 , 5\/16<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>A quadratic polynomial formed for the given sum and product of zeros is given by:<\/p>\n\n\n\n<p>f(x) = x<sup>2<\/sup>&nbsp;+ -(sum of zeros) x + (product of roots)<\/p>\n\n\n\n<p>Here, the sum of zeros is = 21\/8 and product of zero = 5\/16<\/p>\n\n\n\n<p>Thus,<\/p>\n\n\n\n<p>The required polynomial f(x) is,<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;\u2013 (21\/8)x + (5\/16)<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;\u2013 21\/8x + 5\/16<\/p>\n\n\n\n<p>So, to find the zeros we put f(x) = 0<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;\u2013 21\/8x + 5\/16 = 0<\/p>\n\n\n\n<p>\u21d2 16x<sup>2<\/sup>&nbsp;\u2013 42x + 5 = 0<\/p>\n\n\n\n<p>\u21d2 16x<sup>2<\/sup>&nbsp;\u2013 40x \u2013 2x + 5 = 0<\/p>\n\n\n\n<p>\u21d2 8x(2x \u2013 5) \u2013 1(2x \u2013 5) = 0<\/p>\n\n\n\n<p>\u21d2 (2x \u2013 5) (8x \u2013 1) = 0<\/p>\n\n\n\n<p>\u21d2 (2x \u2013 5) = 0 and, or (8x \u2013 1) = 0<\/p>\n\n\n\n<p>Therefore, the two zeros are 5\/2 and 1\/8.<\/p>\n\n\n\n<p><strong>(iii) -2\u221a3, -9<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>A quadratic polynomial formed for the given sum and product of zeros is given by:<\/p>\n\n\n\n<p>f(x) = x<sup>2<\/sup>&nbsp;+ -(sum of zeros) x + (product of roots)<\/p>\n\n\n\n<p>Here, the sum of zeros is = -2\u221a3 and product of zero = -9<\/p>\n\n\n\n<p>Thus,<\/p>\n\n\n\n<p>The required polynomial f(x) is,<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;\u2013 (-2\u221a3)x + (-9)<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;+ 2\u221a3x \u2013 9<\/p>\n\n\n\n<p>So, to find the zeros we put f(x) = 0<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;+ 2\u221a3x \u2013 9 = 0<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;+ 3\u221a3x \u2013 \u221a3x \u2013 9 = 0<\/p>\n\n\n\n<p>\u21d2 x(x + 3\u221a3) \u2013 \u221a3(x + 3\u221a3) = 0<\/p>\n\n\n\n<p>\u21d2 (x + 3\u221a3) (x \u2013 \u221a3) = 0<\/p>\n\n\n\n<p>\u21d2 (x + 3\u221a3) = 0 and, or (x \u2013 \u221a3) = 0<\/p>\n\n\n\n<p>Therefore, the two zeros are -3\u221a3and \u221a3.<\/p>\n\n\n\n<p><strong>(iv) -3\/2\u221a5, -1\/2<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>A quadratic polynomial formed for the given sum and product of zeros is given by:<\/p>\n\n\n\n<p>f(x) = x<sup>2<\/sup>&nbsp;+ -(sum of zeros) x + (product of roots)<\/p>\n\n\n\n<p>Here, the sum of zeros is = -3\/2\u221a5 and product of zero = -1\/2<\/p>\n\n\n\n<p>Thus,<\/p>\n\n\n\n<p>The required polynomial f(x) is,<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;\u2013 (-3\/2\u221a5)x + (-1\/2)<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;+ 3\/2\u221a5x \u2013 1\/2<\/p>\n\n\n\n<p>So, to find the zeros we put f(x) = 0<\/p>\n\n\n\n<p>\u21d2 x<sup>2<\/sup>&nbsp;+ 3\/2\u221a5x \u2013 1\/2 = 0<\/p>\n\n\n\n<p>\u21d2 2\u221a5x<sup>2<\/sup>&nbsp;+ 3x \u2013 \u221a5 = 0<\/p>\n\n\n\n<p>\u21d2 2\u221a5x<sup>2<\/sup>&nbsp;+ 5x \u2013 2x \u2013 \u221a5 = 0<\/p>\n\n\n\n<p>\u21d2 \u221a5x(2x + \u221a5) \u2013 1(2x + \u221a5) = 0<\/p>\n\n\n\n<p>\u21d2 (2x + \u221a5) (\u221a5x \u2013 1) = 0<\/p>\n\n\n\n<p>\u21d2 (2x + \u221a5) = 0 and, or (\u221a5x \u2013 1) = 0<\/p>\n\n\n\n<p>Therefore, the two zeros are -\u221a5\/2 and 1\/\u221a5.<\/p>\n\n\n\n<p><strong>3.<\/strong>&nbsp;<strong>If&nbsp;\u03b1 and \u03b2&nbsp;are the zeros of the quadratic polynomial&nbsp;f(x) = x<sup>2&nbsp;<\/sup>\u2013 5x + 4, find the value of&nbsp;1\/\u03b1 + 1\/\u03b2 \u2013 2\u03b1\u03b2.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>From the question, it\u2019s given that:<\/p>\n\n\n\n<p>\u03b1 and \u03b2&nbsp;are the roots of the quadratic polynomial f(x) where a = 1, b = -5 and c = 4<\/p>\n\n\n\n<p>So, we can find<\/p>\n\n\n\n<p>Sum of the roots =&nbsp;\u03b1+\u03b2&nbsp;= -b\/a = \u2013 (-5)\/1 = 5<\/p>\n\n\n\n<p>Product of the roots =&nbsp;\u03b1\u03b2&nbsp;= c\/a = 4\/1 = 4<\/p>\n\n\n\n<p>To find, 1\/\u03b1 +1\/\u03b2 \u2013 2\u03b1\u03b2<\/p>\n\n\n\n<p>\u21d2 [(\u03b1 +\u03b2)\/ \u03b1\u03b2] \u2013 2\u03b1\u03b2<\/p>\n\n\n\n<p>\u21d2 (5)\/ 4 \u2013 2(4) = 5\/4 \u2013 8 = -27\/ 4<\/p>\n\n\n\n<p><strong>4.<\/strong>&nbsp;<strong>If&nbsp;\u03b1 and \u03b2&nbsp;are the zeros of the quadratic polynomial&nbsp;p(y) = 5y<sup>2&nbsp;<\/sup>\u2013 7y + 1, find the value of&nbsp;1\/\u03b1+1\/\u03b2.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>From the question, it\u2019s given that:<\/p>\n\n\n\n<p>\u03b1 and \u03b2&nbsp;are the roots of the quadratic polynomial f(x) where a =5, b = -7 and c = 1<\/p>\n\n\n\n<p>So, we can find<\/p>\n\n\n\n<p>Sum of the roots =&nbsp;\u03b1+\u03b2&nbsp;= -b\/a = \u2013 (-7)\/5 = 7\/5<\/p>\n\n\n\n<p>Product of the roots =&nbsp;\u03b1\u03b2&nbsp;= c\/a = 1\/5<\/p>\n\n\n\n<p>To find, 1\/\u03b1 +1\/\u03b2<\/p>\n\n\n\n<p>\u21d2 (\u03b1 +\u03b2)\/ \u03b1\u03b2<\/p>\n\n\n\n<p>\u21d2 (7\/5)\/ (1\/5) = 7<\/p>\n\n\n\n<p><strong>5.<\/strong>&nbsp;<strong>If&nbsp;\u03b1 and \u03b2&nbsp;are the zeros of the quadratic polynomial&nbsp;f(x)=x<sup>2<\/sup>&nbsp;\u2013 x \u2013 4, find the value of&nbsp;1\/\u03b1+1\/\u03b2\u2013\u03b1\u03b2.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>From the question, it\u2019s given that:<\/p>\n\n\n\n<p>\u03b1 and \u03b2&nbsp;are the roots of the quadratic polynomial f(x) where a = 1, b = -1 and c = \u2013 4<\/p>\n\n\n\n<p>So, we can find<\/p>\n\n\n\n<p>Sum of the roots =&nbsp;\u03b1+\u03b2&nbsp;= -b\/a = \u2013 (-1)\/1 = 1<\/p>\n\n\n\n<p>Product of the roots =&nbsp;\u03b1\u03b2&nbsp;= c\/a = -4 \/1 = \u2013 4<\/p>\n\n\n\n<p>To find, 1\/\u03b1 +1\/\u03b2 \u2013 \u03b1\u03b2<\/p>\n\n\n\n<p>\u21d2 [(\u03b1 +\u03b2)\/ \u03b1\u03b2] \u2013 \u03b1\u03b2<\/p>\n\n\n\n<p>\u21d2 [(1)\/ (-4)] \u2013 (-4) = -1\/4 + 4 = 15\/ 4<\/p>\n\n\n\n<p><strong>6<\/strong>.&nbsp;<strong>If&nbsp;\u03b1 and \u03b2&nbsp;are the zeroes of the quadratic polynomial&nbsp;f(x) = x<sup>2&nbsp;<\/sup>+ x \u2013 2, find the value of&nbsp;1\/\u03b1 \u2013 1\/\u03b2.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>From the question, it\u2019s given that:<\/p>\n\n\n\n<p>\u03b1 and \u03b2&nbsp;are the roots of the quadratic polynomial f(x) where a = 1, b = 1 and c = \u2013 2<\/p>\n\n\n\n<p>So, we can find<\/p>\n\n\n\n<p>Sum of the roots =&nbsp;\u03b1+\u03b2&nbsp;= -b\/a = \u2013 (1)\/1 = -1<\/p>\n\n\n\n<p>Product of the roots =&nbsp;\u03b1\u03b2&nbsp;= c\/a = -2 \/1 = \u2013 2<\/p>\n\n\n\n<p>To find, 1\/\u03b1 \u2013 1\/\u03b2<\/p>\n\n\n\n<p>\u21d2 [(\u03b2 \u2013 \u03b1)\/ \u03b1\u03b2]<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"381\" height=\"51\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2.png\" alt=\"\" class=\"wp-image-544450\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2.png 381w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-300x40.png 300w\" sizes=\"auto, (max-width: 381px) 100vw, 381px\" \/><\/figure>\n\n\n\n<p>\u21d2<\/p>\n\n\n\n<p><strong>7.<\/strong>&nbsp;<strong>If one of the zero of the quadratic polynomial&nbsp;f(x) = 4x<sup>2&nbsp;<\/sup>\u2013 8kx \u2013 9&nbsp;is negative of the other, then find the value of k.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>From the question, it\u2019s given that:<\/p>\n\n\n\n<p>The quadratic polynomial f(x) where a = 4, b = -8k and c = \u2013 9<\/p>\n\n\n\n<p>And, for roots to be negative of each other, let the roots be \u03b1 and \u2013 \u03b1.<\/p>\n\n\n\n<p>So, we can find<\/p>\n\n\n\n<p>Sum of the roots =&nbsp;\u03b1 \u2013 \u03b1&nbsp;= -b\/a = \u2013 (-8k)\/1 = 8k = 0 [\u2235 \u03b1 \u2013 \u03b1&nbsp;= 0]<\/p>\n\n\n\n<p>\u21d2 k = 0<\/p>\n\n\n\n<p><strong>8.<\/strong>&nbsp;<strong>&nbsp;If the sum of the zeroes of the quadratic polynomial&nbsp;f(t)=kt<sup>2&nbsp;<\/sup>+ 2t + 3k&nbsp;is equal to their product, then find the value of k.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>The quadratic polynomial f(t)=kt<sup>2&nbsp;<\/sup>+ 2t + 3k,<strong>&nbsp;<\/strong>where a = k, b = 2 and c = 3k.<\/p>\n\n\n\n<p>And,<\/p>\n\n\n\n<p>Sum of the roots = Product of the roots<\/p>\n\n\n\n<p>\u21d2 (-b\/a) = (c\/a)<\/p>\n\n\n\n<p>\u21d2 (-2\/k) = (3k\/k)<\/p>\n\n\n\n<p>\u21d2 (-2\/k) = 3<\/p>\n\n\n\n<p>\u2234 k = -2\/3<\/p>\n\n\n\n<p><strong>9.<\/strong>&nbsp;<strong>If&nbsp;\u03b1 and \u03b2&nbsp;are the zeros of the quadratic polynomial&nbsp;p(x) = 4x<sup>2&nbsp;<\/sup>\u2013 5x \u2013 1, find the value of&nbsp;\u03b1<sup>2<\/sup>\u03b2+\u03b1\u03b2<sup>2<\/sup>.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>From the question, it\u2019s given that:<\/p>\n\n\n\n<p>\u03b1 and \u03b2&nbsp;are the roots of the quadratic polynomial p(x) where a = 4, b = -5 and c = -1<\/p>\n\n\n\n<p>So, we can find<\/p>\n\n\n\n<p>Sum of the roots =&nbsp;\u03b1+\u03b2&nbsp;= -b\/a = \u2013 (-5)\/4 = 5\/4<\/p>\n\n\n\n<p>Product of the roots =&nbsp;\u03b1\u03b2&nbsp;= c\/a = -1\/4<\/p>\n\n\n\n<p>To find, \u03b1<sup>2<\/sup>\u03b2+\u03b1\u03b2<sup>2<\/sup><\/p>\n\n\n\n<p>\u21d2 \u03b1\u03b2(\u03b1 +\u03b2)<\/p>\n\n\n\n<p>\u21d2 (-1\/4)(5\/4) = -5\/16<\/p>\n\n\n\n<p><strong>10<\/strong>.&nbsp;<strong>If&nbsp;\u03b1 and \u03b2&nbsp;are the zeros of the quadratic polynomial&nbsp;f(t)=t<sup>2<\/sup>\u2013 4t + 3, find the value of&nbsp;\u03b1<sup>4<\/sup>\u03b2<sup>3<\/sup>+\u03b1<sup>3<\/sup>\u03b2<sup>4<\/sup>.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>From the question, it\u2019s given that:<\/p>\n\n\n\n<p>\u03b1 and \u03b2&nbsp;are the roots of the quadratic polynomial f(t) where a = 1, b = -4 and c = 3<\/p>\n\n\n\n<p>So, we can find<\/p>\n\n\n\n<p>Sum of the roots =&nbsp;\u03b1+\u03b2&nbsp;= -b\/a = \u2013 (-4)\/1 = 4<\/p>\n\n\n\n<p>Product of the roots =&nbsp;\u03b1\u03b2&nbsp;= c\/a = 3\/1 = 3<\/p>\n\n\n\n<p>To find, \u03b1<sup>4<\/sup>\u03b2<sup>3<\/sup>+\u03b1<sup>3<\/sup>\u03b2<sup>4<\/sup><\/p>\n\n\n\n<p>\u21d2 \u03b1<sup>3<\/sup>\u03b2<sup>3&nbsp;<\/sup>(\u03b1 +\u03b2)<\/p>\n\n\n\n<p>\u21d2 (\u03b1\u03b2)<sup>3&nbsp;<\/sup>(\u03b1 +\u03b2)<\/p>\n\n\n\n<p>\u21d2 (3)<sup>3&nbsp;<\/sup>(4) = 27 x 4 = 108<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 2.2 Page No: 2.43<\/h4>\n\n\n\n<p><strong>1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeros and coefficients in each of the following cases:<\/strong><\/p>\n\n\n\n<p><strong>(i) f(x) = 2x<sup>3&nbsp;<\/sup>+ x<sup>2&nbsp;<\/sup>\u2013 5x + 2; 1\/2, 1, -2<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given, f(x) = 2x<sup>3&nbsp;<\/sup>+ x<sup>2&nbsp;<\/sup>\u2013 5x + 2, where a= 2, b= 1, c= -5 and d= 2<\/p>\n\n\n\n<p>For x = 1\/2<\/p>\n\n\n\n<p>f(1\/2) = 2(1\/2)<sup>3<\/sup>&nbsp;+ (1\/2)<sup>2<\/sup>&nbsp;\u2013 5(1\/2) + 2<\/p>\n\n\n\n<p>= 1\/4 + 1\/4 \u2013 5\/2 + 2 = 0<\/p>\n\n\n\n<p>\u21d2 f(1\/2) = 0, hence x = 1\/2 is a root of the given polynomial.<\/p>\n\n\n\n<p>For x = 1<\/p>\n\n\n\n<p>f(1) = 2(1)<sup>3<\/sup>&nbsp;+ (1)<sup>2<\/sup>&nbsp;\u2013 5(1) + 2<\/p>\n\n\n\n<p>= 2 + 1 \u2013 5 + 2 = 0<\/p>\n\n\n\n<p>\u21d2 f(1) = 0, hence x = 1 is also a root of the given polynomial.<\/p>\n\n\n\n<p>For x = -2<\/p>\n\n\n\n<p>f(-2) = 2(-2)<sup>3<\/sup>&nbsp;+ (-2)<sup>2<\/sup>&nbsp;\u2013 5(-2) + 2<\/p>\n\n\n\n<p>= -16 + 4 + 10 + 2 = 0<\/p>\n\n\n\n<p>\u21d2 f(-2) = 0, hence x = -2 is also a root of the given polynomial.<\/p>\n\n\n\n<p>Now,<\/p>\n\n\n\n<p>Sum of zeros = -b\/a<\/p>\n\n\n\n<p>1\/2 + 1 \u2013 2 = \u2013 (1)\/2<\/p>\n\n\n\n<p>-1\/2 = -1\/2<\/p>\n\n\n\n<p>Sum of the products of the zeros taken two at a time = c\/a<\/p>\n\n\n\n<p>(1\/2 x 1) + (1 x -2) + (1\/2 x -2) = -5\/ 2<\/p>\n\n\n\n<p>1\/2 \u2013 2 + (-1) = -5\/2<\/p>\n\n\n\n<p>-5\/2 = -5\/2<\/p>\n\n\n\n<p>Product of zeros = \u2013 d\/a<\/p>\n\n\n\n<p>1\/2 x 1 x (\u2013 2) = -(2)\/2<\/p>\n\n\n\n<p>-1 = -1<\/p>\n\n\n\n<p>Hence, the relationship between the zeros and coefficients is verified.<\/p>\n\n\n\n<p><strong>(ii) g(x) = x<sup>3&nbsp;<\/sup>\u2013 4x<sup>2&nbsp;<\/sup>+ 5x \u2013 2; 2, 1, 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given, g(x) = x<sup>3&nbsp;<\/sup>\u2013 4x<sup>2&nbsp;<\/sup>+ 5x \u2013 2, where a= 1, b= -4, c= 5 and d= -2<\/p>\n\n\n\n<p>For x = 2<\/p>\n\n\n\n<p>g(2) = (2)<sup>3<\/sup>&nbsp;\u2013 4(2)<sup>2<\/sup>&nbsp;+ 5(2) \u2013 2<\/p>\n\n\n\n<p>= 8 \u2013 16 + 10 \u2013 2 = 0<\/p>\n\n\n\n<p>\u21d2 f(2) = 0, hence x = 2 is a root of the given polynomial.<\/p>\n\n\n\n<p>For x = 1<\/p>\n\n\n\n<p>g(1) = (1)<sup>3<\/sup>&nbsp;\u2013 4(1)<sup>2<\/sup>&nbsp;+ 5(1) \u2013 2<\/p>\n\n\n\n<p>= 1 \u2013 4 + 5 \u2013 2 = 0<\/p>\n\n\n\n<p>\u21d2 g(1) = 0, hence x = 1 is also a root of the given polynomial.<\/p>\n\n\n\n<p>Now,<\/p>\n\n\n\n<p>Sum of zeros = -b\/a<\/p>\n\n\n\n<p>1 + 1 + 2 = \u2013 (-4)\/1<\/p>\n\n\n\n<p>4 = 4<\/p>\n\n\n\n<p>Sum of the products of the zeros taken two at a time = c\/a<\/p>\n\n\n\n<p>(1 x 1) + (1 x 2) + (2 x 1) = 5\/ 1<\/p>\n\n\n\n<p>1 + 2 + 2 = 5<\/p>\n\n\n\n<p>5 = 5<\/p>\n\n\n\n<p>Product of zeros = \u2013 d\/a<\/p>\n\n\n\n<p>1 x 1 x 2 = -(-2)\/1<\/p>\n\n\n\n<p>2 = 2<\/p>\n\n\n\n<p>Hence, the relationship between the zeros and coefficients is verified.<\/p>\n\n\n\n<p><strong>2.<\/strong>&nbsp;<strong>Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and product of its zeros as 3, -1 and -3 respectively.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Generally,<\/strong><\/p>\n\n\n\n<p>A cubic polynomial say, f(x) is of the form ax<sup>3&nbsp;<\/sup>+ bx<sup>2&nbsp;<\/sup>+ cx + d.<\/p>\n\n\n\n<p>And, can be shown w.r.t its relationship between roots as.<\/p>\n\n\n\n<p>\u21d2 f(x) = k [x<sup>3<\/sup>&nbsp;\u2013 (sum of roots)x<sup>2<\/sup>&nbsp;+ (sum of products of roots taken two at a time)x \u2013 (product of roots)]<\/p>\n\n\n\n<p>Where, k is any non-zero real number.<\/p>\n\n\n\n<p>Here,<\/p>\n\n\n\n<p>f(x) = k [x<sup>3<\/sup>&nbsp;\u2013 (3)x<sup>2<\/sup>&nbsp;+ (-1)x \u2013 (-3)]<\/p>\n\n\n\n<p>\u2234 f(x) = k [x<sup>3<\/sup>&nbsp;\u2013 3x<sup>2<\/sup>&nbsp;\u2013 x + 3)]<\/p>\n\n\n\n<p>where, k is any non-zero real number.<\/p>\n\n\n\n<p><strong>3.<\/strong>&nbsp;<strong>If the zeros of the polynomial f(x) = 2x<sup>3<\/sup>&nbsp;\u2013 15x<sup>2<\/sup>&nbsp;+ 37x \u2013 30 are in A.P., find them.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Let the zeros of the given polynomial be \u03b1, \u03b2 and \u03b3. (3 zeros as it\u2019s a cubic polynomial)<\/p>\n\n\n\n<p><strong>And given, the zeros are in A.P.<\/strong><\/p>\n\n\n\n<p><strong>So, let\u2019s consider the roots as<\/strong><\/p>\n\n\n\n<p>\u03b1 = a \u2013 d, \u03b2 = a and \u03b3 = a +d<\/p>\n\n\n\n<p>Where, a is the first term and d is the common difference.<\/p>\n\n\n\n<p>From given f(x), a= 2, b= -15, c= 37 and d= 30<\/p>\n\n\n\n<p>\u21d2 Sum of roots = \u03b1 + \u03b2 + \u03b3 = (a \u2013 d) + a + (a + d) = 3a = (-b\/a) = -(-15\/2) = 15\/2<\/p>\n\n\n\n<p>So, calculating for a, we get 3a = 15\/2 \u21d2 a = 5\/2<\/p>\n\n\n\n<p>\u21d2 Product of roots = (a \u2013 d) x (a) x (a + d) = a(a<sup>2<\/sup>&nbsp;\u2013d<sup>2<\/sup>) = -d\/a = -(30)\/2 = 15<\/p>\n\n\n\n<p>\u21d2 a(a<sup>2<\/sup>&nbsp;\u2013d<sup>2<\/sup>) = 15<\/p>\n\n\n\n<p>Substituting the value of a, we get<\/p>\n\n\n\n<p>\u21d2 (5\/2)[(5\/2)<sup>2<\/sup>&nbsp;\u2013d<sup>2<\/sup>] = 15<\/p>\n\n\n\n<p>\u21d2 5[(25\/4) \u2013d<sup>2<\/sup>] = 30<\/p>\n\n\n\n<p>\u21d2 (25\/4) \u2013 d<sup>2<\/sup>&nbsp;= 6<\/p>\n\n\n\n<p>\u21d2 25 \u2013 4d<sup>2<\/sup>&nbsp;= 24<\/p>\n\n\n\n<p>\u21d2 1 = 4d<sup>2<\/sup><\/p>\n\n\n\n<p>\u2234 d = 1\/2 or -1\/2<\/p>\n\n\n\n<p>Taking d = 1\/2 and a = 5\/2<\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>the zeros as 2, 5\/2 and 3<\/p>\n\n\n\n<p>Taking d = -1\/2 and a = 5\/2<\/p>\n\n\n\n<p>We get,<\/p>\n\n\n\n<p>the zeros as 3, 5\/2 and 2<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Exercise 2.3 Page No: 2.57<\/h3>\n\n\n\n<p><strong>1. Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;f(x) = x<sup>3<\/sup>\u2013 6x<sup>2&nbsp;<\/sup>+ 11x \u2013 6, g(x) = x<sup>2&nbsp;<\/sup>+ x +1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>f(x) = x<sup>3<\/sup>\u2013 6x<sup>2&nbsp;<\/sup>+11x \u2013 6, g(x) = x<sup>2&nbsp;<\/sup>+x + 1<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"367\" height=\"251\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-1.png\" alt=\"\" class=\"wp-image-544452\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 1\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-1.png 367w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-1-300x205.png 300w\" sizes=\"auto, (max-width: 367px) 100vw, 367px\" \/><\/figure>\n\n\n\n<p>Thus,<\/p>\n\n\n\n<p>q(x) = x \u2013 7 and r(x) = 17x +1<\/p>\n\n\n\n<p><strong>(ii) f(x) =&nbsp;10x<sup>4&nbsp;<\/sup>+ 17x<sup>3&nbsp;<\/sup>\u2013 62x<sup>2&nbsp;<\/sup>+ 30x \u2013 3, g(x) = 2x<sup>2<\/sup>&nbsp;+ 7x + 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Given,<\/strong><\/p>\n\n\n\n<p><strong>f(x) =&nbsp;<\/strong>10x<sup>4&nbsp;<\/sup>+ 17x<sup>3&nbsp;<\/sup>\u2013 62x<sup>2&nbsp;<\/sup>+ 30x \u2013 3 and g(x) = 2x<sup>2<\/sup><strong>&nbsp;+ 7x + 1<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"549\" height=\"377\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-2.png\" alt=\"\" class=\"wp-image-544453\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 2\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-2.png 549w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-2-300x206.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-2-400x275.png 400w\" sizes=\"auto, (max-width: 549px) 100vw, 549px\" \/><\/figure>\n\n\n\n<p><strong>Thus,<\/strong><\/p>\n\n\n\n<p>q(x) = 5x<sup>2<\/sup>&nbsp;\u2013 9x \u2013 2 and r(x) = 53x \u2013 1<\/p>\n\n\n\n<p><strong>(iii)&nbsp;f(x) = 4x<sup>3<\/sup>&nbsp;+ 8x<sup>2<\/sup>&nbsp;+ 8x + 7, g(x)= 2x<sup>2&nbsp;<\/sup>\u2013 x + 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>f(x) = 4x<sup>3<\/sup>&nbsp;+ 8x<sup>2<\/sup>&nbsp;+ 8x + 7 and g(x)= 2x<sup>2&nbsp;<\/sup>\u2013 x + 1<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"396\" height=\"264\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-3.png\" alt=\"\" class=\"wp-image-544454\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 3\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-3.png 396w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-3-300x200.png 300w\" sizes=\"auto, (max-width: 396px) 100vw, 396px\" \/><\/figure>\n\n\n\n<p><strong>Thus,<\/strong><\/p>\n\n\n\n<p>q(x) = 2x + 5 and r(x) = 11x + 2<\/p>\n\n\n\n<p><strong>(iv)&nbsp;f(x) = 15x<sup>3&nbsp;<\/sup>\u2013 20x<sup>2<\/sup>&nbsp;+ 13x \u2013 12, g(x) = x<sup>2&nbsp;<\/sup>\u2013 2x + 2<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>f(x) = 15x<sup>3&nbsp;<\/sup>\u2013 20x<sup>2<\/sup>&nbsp;+ 13x \u2013 12 and g(x) = x<sup>2&nbsp;<\/sup>\u2013 2x + 2<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"439\" height=\"268\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-4.png\" alt=\"\" class=\"wp-image-544455\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 4\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-4.png 439w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-4-300x183.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-4-400x244.png 400w\" sizes=\"auto, (max-width: 439px) 100vw, 439px\" \/><\/figure>\n\n\n\n<p>Thus,<\/p>\n\n\n\n<p>q(x) = 15x + 10 and r(x) = 3x \u2013 32<\/p>\n\n\n\n<p><strong>2. Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:<\/strong><\/p>\n\n\n\n<p><strong>(i)&nbsp;g(t) = t<sup>2<\/sup>\u20133; f(t)=2t<sup>4<\/sup>&nbsp;+ 3t<sup>3<\/sup>&nbsp;\u2013 2t<sup>2<\/sup>&nbsp;\u2013 9t \u2013 12<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>g(t) = t<sup>2&nbsp;<\/sup>\u2013 3; f(t) =2t<sup>4<\/sup>&nbsp;+ 3t<sup>3<\/sup>&nbsp;\u2013 2t<sup>2<\/sup>&nbsp;\u2013 9t \u2013 12<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"413\" height=\"369\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-5.png\" alt=\"\" class=\"wp-image-544456\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 5\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-5.png 413w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-5-300x268.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-5-400x357.png 400w\" sizes=\"auto, (max-width: 413px) 100vw, 413px\" \/><\/figure>\n\n\n\n<p>Since, the remainder r(t) = 0 we can say that&nbsp;<strong>the first polynomial is a factor of the second polynomial.<\/strong><\/p>\n\n\n\n<p><strong>(ii)&nbsp;g(x) = x<sup>3<\/sup>&nbsp;\u2013 3x + 1; f(x) = x<sup>5<\/sup>&nbsp;\u2013 4x<sup>3<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;+ 3x + 1<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>g(x) = x<sup>3<\/sup>&nbsp;\u2013 3x + 1; f(x) = x<sup>5<\/sup>&nbsp;\u2013 4x<sup>3<\/sup>&nbsp;+ x<sup>2<\/sup>&nbsp;+ 3x + 1<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"524\" height=\"263\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-6.png\" alt=\"\" class=\"wp-image-544457\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 6\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-6.png 524w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-6-300x151.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-6-400x201.png 400w\" sizes=\"auto, (max-width: 524px) 100vw, 524px\" \/><\/figure>\n\n\n\n<p>Since, the remainder r(x) = 2 and not equal to zero we can say that&nbsp;<strong>the first polynomial is not a factor of the second polynomial.<\/strong><\/p>\n\n\n\n<p><strong>(iii) g(x) = 2x<sup>2<\/sup>\u2013 x + 3; f(x) = 6x<sup>5&nbsp;<\/sup>\u2212 x<sup>4<\/sup>&nbsp;+ 4x<sup>3<\/sup>&nbsp;\u2013 5x<sup>2<\/sup>&nbsp;\u2013 x \u201315<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>g(x) = 2x<sup>2<\/sup>\u2013 x + 3; f(x)=6x<sup>5&nbsp;<\/sup>\u2212 x<sup>4<\/sup>&nbsp;+ 4x<sup>3<\/sup>&nbsp;\u2013 5x<sup>2<\/sup>&nbsp;\u2013 x \u201315<\/p>\n\n\n\n\n\n<p>Since, the remainder r(x) = 0 we can say that&nbsp;<strong>the first polynomial is not a factor of the second polynomial.<\/strong><\/p>\n\n\n\n<p><strong>3.<\/strong>&nbsp;<strong>Obtain all zeroes of the polynomial f(x)= 2x<sup>4&nbsp;<\/sup>+ x<sup>3&nbsp;<\/sup>\u2013 14x<sup>2&nbsp;<\/sup>\u2013 19x\u20136, if two of its zeroes are -2 and -1.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p>f(x)= 2x<sup>4&nbsp;<\/sup>+ x<sup>3&nbsp;<\/sup>\u2013 14x<sup>2&nbsp;<\/sup>\u2013 19x \u2013 6<\/p>\n\n\n\n<p>If the two zeros of the polynomial are -2 and -1, then its factors are (x + 2) and (x + 1)<\/p>\n\n\n\n<p>\u21d2 (x+2)(x+1) = x<sup>2&nbsp;<\/sup>+ x + 2x + 2 = x<sup>2&nbsp;<\/sup>+ 3x +2 \u2026\u2026 (i)<\/p>\n\n\n\n<p>This means that (i) is a factor of f(x). So, performing division algorithm we get,<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"507\" height=\"384\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-8.png\" alt=\"\" class=\"wp-image-544458\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 8\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-8.png 507w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-8-300x227.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-8-200x150.png 200w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-8-400x303.png 400w\" sizes=\"auto, (max-width: 507px) 100vw, 507px\" \/><\/figure>\n\n\n\n<p>The quotient is 2x<sup>2<\/sup>&nbsp;\u2013 5x \u2013 3.<\/p>\n\n\n\n<p>\u21d2 f(x)= (2x<sup>2<\/sup>&nbsp;\u2013 5x \u2013 3)( x<sup>2&nbsp;<\/sup>+ 3x +2)<\/p>\n\n\n\n<p>For obtaining the other 2 zeros of the polynomial<\/p>\n\n\n\n<p>We put,<\/p>\n\n\n\n<p>2x<sup>2<\/sup>&nbsp;\u2013 5x \u2013 3 = 0<\/p>\n\n\n\n<p>\u21d2 (2x + 1)(x \u2013 3) = 0<\/p>\n\n\n\n<p>\u2234 x = -1\/2 or 3<\/p>\n\n\n\n<p>Hence, all the zeros of the polynomial are -2, -1, -1\/2 and 3.<\/p>\n\n\n\n<p><strong>4.<\/strong>&nbsp;<strong>&nbsp;Obtain all zeroes of f(x) = x<sup>3&nbsp;<\/sup>+ 13x<sup>2&nbsp;<\/sup>+ 32x + 20, if one of its zeros is -2.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given,<\/p>\n\n\n\n<p><strong>f(x)=&nbsp;<\/strong>x<sup>3&nbsp;<\/sup>+ 13x<sup>2&nbsp;<\/sup>+ 32x + 20<\/p>\n\n\n\n<p>And, -2 is one of the zeros. So, (x + 2) is a factor of f(x),<\/p>\n\n\n\n<p>Performing division algorithm, we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"356\" height=\"356\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-9.png\" alt=\"\" class=\"wp-image-544459\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 9\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-9.png 356w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-9-300x300.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-9-150x150.png 150w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-9-200x200.png 200w\" sizes=\"auto, (max-width: 356px) 100vw, 356px\" \/><\/figure>\n\n\n\n<p>\u21d2 f(x)= (x<sup>2<\/sup>&nbsp;+ 11x + 10)( x + 2)<\/p>\n\n\n\n<p>So, putting x<sup>2<\/sup>&nbsp;+ 11x + 10 = 0 we can get the other 2 zeros.<\/p>\n\n\n\n<p>\u21d2 (x + 10)(x + 1) = 0<\/p>\n\n\n\n<p>\u2234 x = -10 or -1<\/p>\n\n\n\n<p>Hence, all the zeros of the polynomial are -10, -2 and -1.<\/p>\n\n\n\n<p><strong>5.<\/strong>&nbsp;<strong>Obtain all zeroes of the polynomial&nbsp;f(x) = x<sup>4<\/sup>&nbsp;\u2013 3x<sup>3<\/sup>&nbsp;\u2013 x<sup>2<\/sup>&nbsp;+ 9x \u2013 6, if the two of its zeroes are&nbsp;\u2212\u221a3 and \u221a3.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Given,<\/strong><\/p>\n\n\n\n<p>f(x) = x<sup>4<\/sup>&nbsp;\u2013 3x<sup>3<\/sup>&nbsp;\u2013 x<sup>2<\/sup>&nbsp;+ 9x \u2013 6<\/p>\n\n\n\n<p>Since, two of the zeroes of polynomial are&nbsp;\u2212\u221a3 and \u221a3&nbsp;so,&nbsp;(x + \u221a3) and (x\u2013\u221a3)&nbsp;are factors of f(x).<\/p>\n\n\n\n<p>\u21d2 x<sup>2&nbsp;<\/sup>\u2013 3 is a factor of f(x). Hence, performing division algorithm, we get<br><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"420\" height=\"365\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-10.png\" alt=\"\" class=\"wp-image-544460\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-10.png 420w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-10-300x261.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-10-400x348.png 400w\" sizes=\"auto, (max-width: 420px) 100vw, 420px\" \/><\/figure>\n\n\n\n<p>\u21d2 f(x)= (x<sup>2<\/sup>&nbsp;\u2013 3x + 2)( x<sup>2&nbsp;<\/sup>\u2013 3)<\/p>\n\n\n\n<p>So, putting x<sup>2<\/sup>&nbsp;\u2013 3x + 2 = 0 we can get the other 2 zeros.<\/p>\n\n\n\n<p>\u21d2 (x \u2013 2)(x \u2013 1) = 0<\/p>\n\n\n\n<p>\u2234 x = 2 or 1<\/p>\n\n\n\n<p>Hence, all the zeros of the polynomial are \u2212\u221a3, 1, \u221a3 and 2.<\/p>\n\n\n\n<p><strong>6.<\/strong>&nbsp;<strong>&nbsp;Obtain all zeroes of the polynomial&nbsp;f(x)= 2x<sup>4&nbsp;<\/sup>\u2013 2x<sup>3&nbsp;<\/sup>\u2013 7x<sup>2&nbsp;<\/sup>+ 3x + 6, if the two of its zeroes are&nbsp;\u2212\u221a(3\/2) and \u221a(3\/2).<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Given,<\/strong><\/p>\n\n\n\n<p>f(x)= 2x<sup>4&nbsp;<\/sup>\u2013 2x<sup>3&nbsp;<\/sup>\u2013 7x<sup>2&nbsp;<\/sup>+ 3x + 6<\/p>\n\n\n\n<p>Since, two of the zeroes of polynomial are&nbsp;\u2212\u221a(3\/2) and \u221a(3\/2)&nbsp;so,&nbsp;(x + \u221a(3\/2)) and (x \u2013\u221a(3\/2))&nbsp;are factors of f(x).<\/p>\n\n\n\n<p>\u21d2 x<sup>2&nbsp;<\/sup>\u2013 (3\/2) is a factor of f(x). Hence, performing division algorithm, we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"445\" height=\"368\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-11.png\" alt=\"\" class=\"wp-image-544461\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 10\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-11.png 445w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-11-300x248.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-11-400x331.png 400w\" sizes=\"auto, (max-width: 445px) 100vw, 445px\" \/><\/figure>\n\n\n\n<p>\u21d2 f(x)= (2x<sup>2<\/sup>&nbsp;\u2013 2x \u2013 4)( x<sup>2&nbsp;<\/sup>\u2013 3\/2)= 2(x<sup>2<\/sup>&nbsp;\u2013 x \u2013 2)( x<sup>2&nbsp;<\/sup>\u2013 3\/2)<\/p>\n\n\n\n<p>So, putting x<sup>2<\/sup>&nbsp;\u2013 x \u2013 2 = 0 we can get the other 2 zeros.<\/p>\n\n\n\n<p>\u21d2 (x \u2013 2)(x + 1) = 0<\/p>\n\n\n\n<p>\u2234 x = 2 or -1<\/p>\n\n\n\n<p>Hence, all the zeros of the polynomial are \u2212\u221a(3\/2), -1, \u221a(3\/2) and 2.<\/p>\n\n\n\n<p><strong>7.<\/strong>&nbsp;<strong>&nbsp;Find all the zeroes of the polynomial&nbsp;x<sup>4&nbsp;<\/sup>+ x<sup>3&nbsp;<\/sup>\u2013 34x<sup>2&nbsp;<\/sup>\u2013 4x + 120, if the two of its zeros are 2 and -2.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p><strong>Let,<\/strong><\/p>\n\n\n\n<p>f(x) = x<sup>4<\/sup>&nbsp;+ x<sup>3<\/sup>&nbsp;\u2013 34x<sup>2<\/sup>&nbsp;\u2013 4x + 120<\/p>\n\n\n\n<p>Since, two of the zeroes of polynomial are&nbsp;\u22122 and 2&nbsp;so,&nbsp;(x + 2) and (x \u2013 2)&nbsp;are factors of f(x).<\/p>\n\n\n\n<p>\u21d2 x<sup>2&nbsp;<\/sup>\u2013 4 is a factor of f(x). Hence, performing division algorithm, we get<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"456\" height=\"359\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-12.png\" alt=\"\" class=\"wp-image-544462\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 2 Polynomials ex 2.3 - 11\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-12.png 456w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-12-300x236.png 300w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-2-12-400x315.png 400w\" sizes=\"auto, (max-width: 456px) 100vw, 456px\" \/><\/figure>\n\n\n\n<p>\u21d2 f(x)= (x<sup>2<\/sup>&nbsp;+ x \u2013 30)( x<sup>2&nbsp;<\/sup>\u2013 4)<\/p>\n\n\n\n<p>So, putting x<sup>2<\/sup>&nbsp;+ x \u2013 30 = 0 we can get the other 2 zeros.<\/p>\n\n\n\n<p>\u21d2 (x + 6)(x \u2013 5) = 0<\/p>\n\n\n\n<p>\u2234 x = -6 or 5<\/p>\n\n\n\n<p>Hence, all the zeros of the polynomial are 5, -2, 2 and -6.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-10-maths-chapter-2-download-pdf\">RD Sharma Solutions for Class 10 Maths Chapter 2:&nbsp;<strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>RD Sharma Solutions for Class 10 Maths Chapter 2\u2013Polynomials<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RD-Sharma-Solutions-for-Class-10-Maths-Chapter-2\u2013Polynomials.pdf\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: RD Sharma Solutions for Class 10 Maths Chapter 2\u2013Polynomials PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Chapterwise RD Sharma Solutions for Class 10&nbsp;Maths :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-1-real-numbers\/\">Chapter 1\u2013Real Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/\">Chapter 2\u2013Polynomials<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-3-pair-of-linear-equations-in-two-variables\/\">Chapter 3\u2013Pair of Linear Equations In Two Variables<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-4-triangles\/\">Chapter 4\u2013Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-5-trigonometric-ratios\/\">Chapter 5\u2013Trigonometric Ratios<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-6-trigonometric-identities\/\">Chapter 6\u2013Trigonometric Identities<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-7-statistics\/\">Chapter 7\u2013Statistics<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-8-quadratic-equations\/\">Chapter 8\u2013Quadratic Equations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-9-arithmetic-progressions\/\">Chapter 9\u2013Arithmetic Progressions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-10-circles\/\">Chapter 10\u2013Circles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-11-constructions\/\">Chapter 11\u2013Constructions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-12-some-applications-of-trigonometry\/\">Chapter 12\u2013Some Applications of Trigonometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-13-probability\/\">Chapter 13\u2013Probability<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-14-co-ordinate-geometry\/\">Chapter 14\u2013Co-ordinate Geometry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-15-areas-related-to-circles\/\">Chapter 15\u2013Areas Related To Circles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-16-surface-areas-and-volumes\/\">Chapter 16\u2013Surface Areas And Volumes<\/a><\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">About RD Sharma<\/h2>\n\n\n\n<p>RD Sharma i<em>sn&#8217;t the kind of author you&#8217;d bump into at lit fests. But his bestselling books have helped many&nbsp;<\/em>CBSE<em>&nbsp;students lose their dread of&nbsp;<\/em>maths<em>. Sunday Times profiles the tutor turned internet star<\/em><br>He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like &#8216;series solution of linear differential equations&#8217;. Meet Dr&nbsp;Ravi Dutt Sharma&nbsp;\u2014&nbsp;mathematics&nbsp;teacher and author of 25 reference books \u2014 whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it&#8217;s only recently that a spoof video turned the tutor into a YouTube star.<\/p>\n\n\n\n<p>R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. &#8220;I like to spend all my time thinking and writing about maths problems. I find it relaxing,&#8221; he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government&#8217;s Guru Nanak Dev Institute of Technology.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Read More<\/h2>\n\n\n\n<ul class=\"wp-block-yoast-seo-related-links\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-11-maths-chapter-11-trigonometric-equations\/\">RD Sharma Solutions for Class 11 Maths Chapter 11\u2013Trigonometric Equations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-2-polynomials\/\">NCERT Solutions for 9th Class Maths :Chapter 2 Polynomials<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-chapter-12-areas-related-to-circles\/\">NCERT Solutions for Maths: Chapter 12 \u2013 Areas Related to Circles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-4-quadratic-equations\/\">NCERT Solutions for Class 10th Mathematics: Chapter 4 Quadratic Equations<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-2-polynomials\/\">NCERT Solutions for Class 10th Mathematics: Chapter 2 Polynomials<\/a><\/li><\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Class 10: Maths Chapter 2 solutions. Complete Class 10 Maths Chapter 2 Notes. RD Sharma Solutions for Class 10 Maths Chapter 2\u2013Polynomials RD Sharma 10th Maths Chapter 2, Class 10 Maths Chapter 2 solutions Exercise 2.1 Page No: 2.33 1. Find the zeros of each of the following quadratic polynomials and verify the relationship between [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":544447,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,24],"tags":[1962],"boards":[],"class_list":["post-544444","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-10","tag-rd-sharma-solutions","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>RD Sharma Solutions for Class 10, maths Chapter 2 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"RD Sharma Solutions for Class 10 Maths Chapter 2\u2013Polynomials | Browse all Class 10 Maths Chapters RD Sharma books - IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"RD Sharma Solutions for Class 10 Maths Chapter 2\u2013Polynomials\" \/>\n<meta property=\"og:description\" content=\"Class 10: Maths Chapter 2 solutions. Complete Class 10 Maths Chapter 2 Notes. RD Sharma Solutions for Class 10 Maths Chapter 2\u2013Polynomials RD Sharma 10th\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2021-10-02T05:02:06+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-10-04T05:37:07+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/i2.wp.com\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/class10m2.png?fit=1200%2C675&ssl=1\" \/>\n\t<meta property=\"og:image:width\" content=\"1200\" \/>\n\t<meta property=\"og:image:height\" content=\"675\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"25 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"RD Sharma Solutions for Class 10 Maths Chapter 2\u2013Polynomials\",\"datePublished\":\"2021-10-02T05:02:06+00:00\",\"dateModified\":\"2021-10-04T05:37:07+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/\"},\"wordCount\":3905,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/class10m2.png\",\"keywords\":[\"RD Sharma Solutions\"],\"articleSection\":[\"Book Solutions\",\"Class 10\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-10-maths-chapter-2-polynomials\/\",\"name\":\"RD Sharma Solutions for Class 10, maths Chapter 2 - 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Complete Class 10 Maths Chapter 2 Notes. 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