NCERT Solutions for Class 10 Maths<\/a> Chapter 2 Polynomials Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 10 Maths Chapter 2 Polynomials<\/h2>\nExercise 2.3<\/h2>\n <\/p>\n
Question 1. \nDivide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder, in each of the following: \n(i) p(x) = x3<\/sup> – 3x2<\/sup> + 5x -3, g(x) = x2<\/sup>-2 \n(ii) p(x) =x4<\/sup> – 3x2<\/sup> + 4x + 5, g(x) = x2<\/sup> + 1 -x \n(iii) p(x) = x4<\/sup> – 5x + 6, g(x) = 2 -x2<\/sup> \nSolution: \n(i) \n \nTherefore, \nquotient = x – 3 and remainder = 7x – 9<\/p>\n(ii) First we arrange the terms of the dividend and the divisor in the decreasing order of their degrees. \n\u2234 p(x) = x4<\/sup>– 3x2<\/sup> + 4x + 5 and g(x) = x2<\/sup> – x + 1 \n \nTherefore, \nquotient = x2 +x-3 and remainder = 8<\/p>\n(iii) First we arrange the terms of the dividend and the divisior in the decreasing order of their degrees. \n\u2234 p(x) = x2<\/sup> + x – 3 and g(x) = – x2<\/sup> + 2 \n \nTherefore, \nquotient = – x\u00b2 – 2 and remainder = – 5x + 10<\/p>\n <\/p>\n
Question 2. \nCheck whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: \n(i) t2 <\/sup>– 3, 2t4<\/sup> + t3<\/sup> – 2t2<\/sup> – 9t – 12 \n(ii) x2<\/sup> + 3x + 1, 3x4 <\/sup>+ 5x3 <\/sup>-7x2 <\/sup>+ 2x + 2 \n(iii) x3<\/sup> – 3x + 1, x5<\/sup> – 4x3<\/sup> + x2<\/sup> + 3x + l \nSolution: \nWe have, \nP(t) = 2t4<\/sup> + 3t3<\/sup> – 2t2<\/sup> – 9t – 12 \nq(x) = t2<\/sup> – 3 \nBy actual division, we have \n \nHere, remainder is zero. \nTherefore, q(x) = t2<\/sup> – 3 is the factor of p(x) = 2t4<\/sup> + 3t3<\/sup> – 2t2<\/sup> – 9t – 12.<\/p>\n(ii) We have, \np(x) = 3x4<\/sup> + 5x3<\/sup> – 7x2<\/sup> + 2x + 2 \nand q(x) = x\u00b2 + 3x + 1 \nBy actual division, we have \n \nHere, remainder is not zero. \nTherefore, q(x) = x\u00b2 – 3x + 1 is not a factor of p(x) = 3x4<\/sup> + 5x3<\/sup> – 7x2<\/sup> + 2x + 2.<\/p>\n(iii) We have, \np(x) = x5<\/sup> – x3<\/sup> + x2<\/sup> + 3x + 1 \nand q(x) = x\u00b3 – 3x + 1 \nBy actual division, we have \n \nHere, remainder is not zero. \nTherefore, q(x) = x\u00b2 – 3x + 1 is not a factor of p(x) = x5<\/sup> – x3<\/sup> + x2<\/sup> + 3x + 1.<\/p>\n <\/p>\n
Question 3. \nObtain all other zeroes of 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5, if two of its zeroes are \\(\\sqrt { \\frac { 5 }{ 3 } }\\) and – \\(\\sqrt { \\frac { 5 }{ 3 } }\\) \nSolution: \nWe have given that two zeroes of polynomial p(x) = 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5 are \\(\\sqrt { \\frac { 5 }{ 3 } }\\) and – \\(\\sqrt { \\frac { 5 }{ 3 } }\\) \n\u2234 (x – \\(\\sqrt { \\frac { 5 }{ 3 } }\\))(x + \\(\\sqrt { \\frac { 5 }{ 3 } }\\)) = x\u00b2 – \\(\\frac { 5 }{ 3 }\\) is a factor is given polynomial p(x). \nNow apply the division algorithm to the given polynomial and x\u00b2 – \\(\\frac { 5 }{ 3 }\\) \n \nSo, 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5 \n \nIf 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5 = 0 \nThen, (x + \\(\\sqrt { \\frac { 5 }{ 3 } }\\))(x – \\(\\sqrt { \\frac { 5 }{ 3 } }\\))(x + 1)(3x + 3) = 0 \n\u2234 x = \\(\\sqrt { \\frac { 5 }{ 3 } }\\) or x = \\(\\sqrt { \\frac { 5 }{ 3 } }\\) Therefore, the zeroes of polynomial p(x) = 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5 are – \\(\\sqrt { \\frac { 5 }{ 3 } }\\), \\(\\sqrt { \\frac { 5 }{ 3 } }\\), -1 and -1.<\/p>\n <\/p>\n
Question 4. \nOn dividing x3<\/sup> – 3x2<\/sup> + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Find g(x). \nSolution: \nWe know that \nDividend = Divisor x Quotiet + Remainder \n\u2234 x\u00b2 – 3x\u00b2 + x + 2 = g(x) x (x – 2) + (- 2x + 4) \nor x3<\/sup>– 3x2<\/sup> + x + 2 = g(x) (x – 2) + (-2x + 4) \nor x3 <\/sup>– 3x2<\/sup> + x + 2 + 2 x- 4 = g(x) x (x-2) \nor x3<\/sup> – 3x2<\/sup> + 3x – 2 = g(x) x (x – 2) \n <\/p>\n <\/p>\n
Question 5. \nGive examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and \n(i) deg p(x) = deg q(x) \n(ii) deg q(x) = deg r(x) \n(iii) deg r(x) = 0 \nSolution: \n(i) deg p(x) = deg q(x) \nPolynomial p(x) = 2x2<\/sup>– 2x + 14; g(x) = 2 \nq(x) = x2<\/sup> – x + 7 r(x) = 0 \nHere, deg p(x) = deg q(x)<\/p>\n(ii) deg q(x) = deg r(x) \nPolynomial p(x) = x3<\/sup>+ x2<\/sup> + x + 1; g(x) = x2<\/sup> – 1, \nq(x) = x + 1, r(x) = 2x + 2<\/p>\n(iii) deg r(x) is 0. \nPolynomial p(x) = x2<\/sup>+ 2x2<\/sup> – x + 2; g(x) = x2<\/sup> – 1, q(x) = x + 1, r(x) = 4<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.3 Question 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder, in each of the following: …<\/p>\n
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n