{"id":25324,"date":"2022-06-05T14:00:55","date_gmt":"2022-06-05T08:30:55","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=25324"},"modified":"2022-05-23T15:50:58","modified_gmt":"2022-05-23T10:20:58","slug":"ncert-solutions-for-class-10-maths-chapter-2-ex-2-3","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-2-ex-2-3\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3"},"content":{"rendered":"

These 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>\n

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials<\/h2>\n

Exercise 2.3<\/h2>\n

\"NCERT<\/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\"NCERT
\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\"NCERT
\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\"NCERT
\nTherefore,
\nquotient = – x\u00b2 – 2 and remainder = – 5x + 10<\/p>\n

\"NCERT<\/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\"NCERT
\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\"NCERT
\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\"NCERT
\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

\"NCERT<\/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\"NCERT
\nSo, 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5
\n\"NCERT
\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

\"NCERT<\/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\"NCERT<\/p>\n

\"NCERT<\/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<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-2-ex-2-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts. 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