NCERT Solutions for Class 9 Maths<\/a> Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2<\/h2>\n Question 1. \nFind the value of the polynomial 5x – 4x2<\/sup> + 3 at \n(i) x = 0 \n(ii) x = -1 \n(iii) x = 2 \nSolution: \n(i) Let P(x) = 5x – 4x2<\/sup> + 3 \nTherefore, P(0) = 5(0) – 4(0)2<\/sup> + 3 \n= 0 – 0 + 3 \n= 3 \nSo, the value of P(x) at x = 0 is 3.<\/p>\n(ii) Let P(x) = 5x – 4x2<\/sup> + 3 \nTherefore, P(-1) = 5(-1) – 4(-1)2<\/sup> + 3 \n= -5 – 4 + 3 \n= -6 \nSo, the value of P(x) at x = -1 is -6.<\/p>\n(iii) Let P(x) = 5x – 4x2<\/sup> + 3 \nTherefore, P(2) = 5(2) – 4(2)2<\/sup> + 3 \n= 10 – 16 + 3 \n= -3 \nSo, the value of P(x) at x = 2 is -3.<\/p>\n <\/p>\n
Question 2. \nFind p(0), p(1) and p(2) for each of the following polynomials: \n(i) p(y) = y2<\/sup> – y + 1 \n(ii) p(t) = 2 + t + 2t2<\/sup> – t3<\/sup> \n(iii) p(x) = x3<\/sup> \n(iv) p(x) = (x – 1) (x + 1) \nSolution: \n(i) We have given \np(y) = y2<\/sup> – y + 1 \nTherefore, the value of polynomial p(y) at y = 0 is \np(0) = 02<\/sup> – 0 + 1 = 1 \nAgain, the value of polynomial p(y) at y = 1 is \np(1) = 12<\/sup> – 1 + 1 = 1 \nAgain, the value of polynomial p(y) at y = 2 is \np(2) = 22<\/sup> – 2 + 1 = 3<\/p>\n(ii) We have given, \np(t) = 2 + t + 2t2<\/sup> – t3<\/sup> \nTherefore, the value of polynomial p(t) at t = 0 is \np(0) = 2 + 0 + 2(0)2<\/sup> – (0)3<\/sup> = 2 \nAgain, the value of polynomial p(t) at t = 1 is \np(1) = 2 + 1 + 2(1)2<\/sup> – (1)3<\/sup> = 4 \nAgain, the value of polynomial p(t) at t = 2 is \np(2) = 2 + 2 + 2(2)2<\/sup> – (2)3<\/sup> = 4<\/p>\n(iii) We have given, \nP(x) = x3<\/sup>, \nTherefore, the value of polynomial p(x) at x = 0 is \np(0) = (0)3<\/sup> = 0 \nAgain the value of polynomial p(x) at x = 1 is \np(1) = (1)3<\/sup> = 1 \nAgain, the value of polynomial p(x) at x = 2 is \nP(2) = (2)3<\/sup> = 8<\/p>\n(iv) We have given \np(x) = (x – 1)(x + 1) \nTherefore, the value of polynomial p(x) at x = 0 is \nP(0) = (0 – 1)(0 + 1) \n= (-1) \u00d7 (+1) \n= -1 \nAgain, the value of polynomial p(x) at x = 1 is \np(1) = (1 – 1) \u00d7 (1 + 1) \n= 0 \u00d7 2 \n= 0 \nAgain, the value of polynomial p(x), at x = 2 is \np(2) = (2 – 1)(2 + 1) \n= 1 \u00d7 3 \n= 3<\/p>\n
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Question 3. \nVerify whether the following are zeros of the polynomial indicated against them: \n(i) p(x) = 3x + 1, x = \\(-\\frac{1}{3}\\) \n(ii) p(x) = 5x – p, x = \\(\\frac{4}{5}\\) \n(iii) p(x) = x2<\/sup> – 1, x = 1, -1 \n(iv) p(x) = (x + 1) (x – 2), x = -1, 2 \n(v) p(x) = x2<\/sup>, x = 0 \n(vi) p(x) = 1x + m, x = \\(-\\frac{m}{1}\\) \n(vii) p(x) = 3x2<\/sup> – 1, x = \\(-\\frac{1}{\\sqrt{3}}, \\frac{2}{\\sqrt{3}}\\) \n(viii) p(x) = 2x + 1, x = \\(\\frac{1}{2}\\) \nSolution: \n(i) We have given that, p(x) = 3x + 1. \nTherefore, the value of polynomial p(x) at x = \\(-\\frac{1}{3}\\) is \n\\(P\\left(-\\frac{1}{3}\\right)=3 \\times \\frac{(-1)}{3}+1\\) = 0 \nyes x = \\(-\\frac{1}{3}\\) is the zero of polynomial p(x).<\/p>\n(ii) We have given that p(x) = 5x – p \nTherefore, the value of polynomial p(x) at x = \\(\\frac{4}{5}\\) is \n\\(P\\left(\\frac{4}{5}\\right)=5 \\times \\frac{4}{5}-\\frac{22}{7}\\) (\u2235 \u03c0 = \\(\\frac {22}{7}\\)) \n= 4 – \\(\\frac{22}{7}\\) \n\\(P\\left(\\frac{4}{5}\\right)=\\frac{6}{7}\\) \nNo, x = \\(\\frac{4}{5}\\) is not the zero of p(x) = 5x – \u03c0.<\/p>\n
(iii) We have given that p(x) = x2<\/sup> – 1 \nTherefore, the value of polynomial p(x) at x = 1 is \np(1) = 12<\/sup> – 1 = 0 \nAgain, the value of polynomial p(x) at x = -1 is \np(-1) = (-1)2<\/sup> – 1 = 0 \nyes x = 1, -1 are the zero of polynomial p(x) = x2<\/sup> – 1<\/p>\n(iv) We have given that \np(x) = (x + 1)(x – 2) \nTherefore, the value of polynomial p(x) at x = -1 is \np(-1) = (-1 + 1)(-1 – 2) \n= 0 \u00d7 (-3) \n= 0 \nAgain, the value of polynomial p(x) at x = 2 is \nP(2) = (2 + 1)(2 – 2) \n= 3 \u00d7 0 \n= 0 \nyes x = -1, 2 are the zero of polynomial p(x) = (x + 1)(x – 2)<\/p>\n
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(v) We have given that p(x) = x2<\/sup> \nTherefore, the value of polynomial p(x) at x = 0 is \np(0) = 02<\/sup> = 0 \nyes x = 0 is the zero of polynomial p(x) = x2<\/sup>.<\/p>\n(vi) We have given that, p(x) = lx + m \nTherefore, the value of polynomial p(x) at x = \\(-\\frac{m}{\\ell}\\) is \n\\(P\\left(-\\frac{m}{\\ell}\\right)=\\ell\\left(-\\frac{m}{\\ell}\\right)+m\\) = 0 \nyes x = \\(-\\frac{m}{\\ell}\\) is the zero of polynomial p(x) = lx + m.<\/p>\n
(vii) We have given that, p(x) = 3x2<\/sup> – 1 \nTherefore, the value of polynomial p(x) at x = \\(-\\frac{1}{\\sqrt{3}}\\) is \n\\(P\\left(-\\frac{1}{\\sqrt{3}}\\right)=3 \\times\\left(-\\frac{1}{\\sqrt{3}}\\right)^{2}-1\\) \n= \\(3 \\times \\frac{1}{3}-1\\) \n= 0 \nAgain, the value of polynomial p(x) at x = \\(\\frac{2}{\\sqrt{3}}\\) is \n\\(P\\left(\\frac{2}{\\sqrt{3}}\\right)=3 \\times\\left(\\frac{2}{\\sqrt{3}}\\right)^{2}-1\\) \n= \\(3 \\times \\frac{4}{3}-1\\) \n= 4 – 1 \n= 3 \nTherefore, x = \\(-\\frac{1}{\\sqrt{3}}\\) is the zero of polynomial p(x) and x = \\(\\frac{2}{\\sqrt{3}}\\) is not the zero of polynomial p(x).<\/p>\n(viii) We have given that p(x) = 2x + 1 \nTherefore, the value of polynomial p(x) at x = \\(\\frac{1}{2}\\) is \n\\(\\mathrm{P}\\left(\\frac{1}{2}\\right)=2 \\times \\frac{1}{2}+1\\) = 2 \nTherefore, x = \\(\\frac{1}{2}\\) is not the zero of polynomial p(x).<\/p>\n
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Question 4. \nFind the zero of the polynomial in each of the following cases: \n(i) p(x) = x + 5 \n(ii) p(x) = x – 5 \n(iii) p(x) = 2x + 5 \n(iv) p(x) = 3x – 2 \n(v) p(x) = 3x \n(vi) p(x) = ax, a \u2260 0 \n(vii) p(x) = cx + d, c \u2260 0; c, d are real numbers. \nSolution: \n(i) We have given that \np(x) = x + 5 ……..(i) \nTo find the zero of polynomial p(x), we can take \nP(x) = 0 ………(ii) \nfrom equation (i) and (ii) \nx + 5 = 0 \n\u2234 x = -5 \nTherefore, x = -5 is the zero of polynomial p(x) = x + 5.<\/p>\n
(ii) We have given that, \np(x) = x – 5 ……..(i) \nTo find the zero of polynomial p(x) we can take \np(x) = 0 ……..(ii) \nfrom equation (i) and (ii) \nx – 5 = 0 \n\u2234 x = 5 \nTherefore x = 5 is the zero of polynomial p(x) = x – 5.<\/p>\n
(iii) We have given that \np(x) = 2x + 5 ………(i) \nTo find the zero of polynomial p(x), we can take \np(x) = 0 …….(ii) \nfrom equation (i) and (ii) \n2x + 5 = 0 \nor x = \\(-\\frac{5}{2}\\) \nTherefore x = \\(-\\frac{5}{2}\\) is the zero of polynomial p(x) = 2x + 5.<\/p>\n
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(iv) We have given that \np(x) = 3x – 2 …….(i) \nTo find the zero of polynomial p(x) we can take \np(x) = 0 ……..(ii) \n3x – 2 = 0 \nx = \\(\\frac{2}{3}\\) \nTherefore, x = \\(\\frac{2}{3}\\) is the zero of polynomial p(x) = 3x – 2.<\/p>\n
(v) We have given that \np(x) = 3x ……..(i) \nTo find the zero of polynomial p(x) we can take \nP(x) = 0 ……..(ii) \nfrom equation (i) and (ii) \n3x = 0 \nx = 0 \nTherefore, x = 0 is the zero of polynomial p(x) = 3x.<\/p>\n
(vi) We have given that, \np(x) = ax, a \u2260 0 ……(i) \nTo find the zero of polynomial p(x) we can take \np(x) = 0 ………(ii) \nfrom equation (i) and (ii) \nax = 0 \nx = \\(\\frac{0}{a}\\) = 0 \nTherefore, x = 0 is the zero of polynomial p(x) = ax, a \u2260 0.<\/p>\n
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(vii) We have given \np(x) = cx + d …….(i) \nc \u2260 0, c, d are real numbers. \nTo find the zero of polynomial p(x) we can take \np(x) = 0 ………(ii) \nfrom equation (i) and (ii) \ncx + d = 0 \nx = \\(-\\frac{d}{c}\\) \nwhere c \u2260 0 and c, d are real numbers. \nTherefore, x = \\(\\frac{d}{c}\\), where c \u2260 0 and c, d are real numbers is the solution of polynomial p(x) = cx + d.<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Question 1. Find the value of the polynomial 5x – 4×2 + 3 at (i) x = 0 (ii) x = …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2<\/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":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 - 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