NCERT Solutions for Class 10 Maths<\/a> Chapter 2 Polynomials Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 10 Maths Chapter 2 Polynomials<\/h2>\nExercise 2.4<\/h2>\n <\/p>\n
Question 1. \nVerify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case: \n(i) 2x3<\/sup> + x2<\/sup> – 5x + 2; \u00a0\\(\\frac { 1 }{ 2 }\\), 1, – 2 \n(ii) x3<\/sup> – 4x2<\/sup> + 5x – 2; 2, 1, 1 \nSolution: \n(i) Comparing the given polynomial with ax3<\/sup> + bx2<\/sup> + cx + d, we get: \na = 2, b – 1, c = -5 and d = 2. \nNow, we get the zeroes \n \nTherefore, \\(\\frac { 1 }{ 2 }\\), 1, – 2 are the zeroes of 2x\u00b3 + x\u00b2 – 5x + 2. \nSo we take \u03b1, \u00df, \u03b3 are the co-efficient of cubic polynomial. \nSum of zeroes \n <\/p>\n(ii) x3<\/sup> – 4x2<\/sup> + 5x – 2; 2, 1, 1 \nCompearing the given polynomial with ax3<\/sup> + bx2<\/sup> + cx + d, we get: \na = 1, b = -4, c = 5 and d = – 2. \n\u2234 p (x) = x3<\/sup> – 4x2<\/sup> + 5x – 2 \n\u21d2\u00a0 p(2) = (2)3<\/sup> – 4(2)2<\/sup> + 5 x 2 – 2 \n= 8 – 16+ 10 – 2 = 0 \np(1) = (1)3<\/sup> – 4(1)2<\/sup> + 5 x 1- 2 \n= 1 – 4 + 1 – 2 \n= 6 – 6 = 0 \nTherefore, 2, 1 and 1 are the zeroes of x3<\/sup> – 4x2 <\/sup>+ 5x – 2. \nSo, take \u03b1, \u00df, \u03b3 are the coefficient of cubic polynomial. \nComparing x\u00b2 – 4x\u00b2 + 5x – 2 = 0 \nSum of zeroes \n <\/p>\n <\/p>\n
Question 2. \nFind a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively. \nSolution: \nWe know that a cubic equation is ax\u00b3 + bx\u00b2 + cx + d = 0 \nBut given, \u03b1 + \u03b2 + \u03b3 = 2 \n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = -7 and \u03b1\u03b2\u03b3 = – 14 \nAlso, we know that a + b + g = \\(\\frac { -b }{ a }\\) = 2, hence, b = – 2 \n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = \\(\\frac { c }{ a }\\) = – 7, hence c = – 7 a \n\u03b1\u03b2\u03b3 = \\(\\frac { -d }{ a }\\) = – 14, hence d = – 14, and a = 1 \nNow, put the value of a, b, c, d in equation (1), we get \nx\u00b3 – 2x\u00b2 – 7x + 14 = 0<\/p>\n
Question 3. \nIf the zeroes of the polynomial x3<\/sup> – 3x2<\/sup> + x + 1 are a-b, a, a + b, find a and b. \nSolution: \nZeroes of the polynomial are a-b, a, a + b \nSum of zeroes = a-b + a + a + b = 3a \n \n\u21d2 a\u00b3 – b\u00b2a = 1 \nPut a = 1, in a\u00b2 – b\u00b2a = – 1 \n(1)\u00b3 – b\u00b2 = – 1 \u21d2 b\u00b2 = 2 \nb= \u00b1 \\(\\sqrt{2}\\) \nvalues of a and b are 1, \u00b1 \\(\\sqrt{2}\\)<\/p>\n <\/p>\n
Question 4. \nIf two zeroes of the polynomial x4<\/sup> – 6x3<\/sup> – 26x2 <\/sup>+ 138x – 35 are 2 \u00b1 \\(\\sqrt{3}\\), finnd other zeroes. \nSolution: \nLet two zeroes are 2 + \\(\\sqrt{3}\\) and 2 – \\(\\sqrt{3}\\), \n \n(x\u00b2 – 4x +1) is a factor of the given polynomial. \nNow, we divide the given polynomial by \n \nSo, x4<\/sup> – 6x3<\/sup> – 26x2<\/sup> + 138x – 35 = (x2<\/sup> – 4x + 1) (x2<\/sup> – 2x – 35) \nNow, by spilitting – 2x, we factorise x\u00b2 – 2x – 35 \n= x3<\/sup> – 7x + 5x – 35 \n= x(x – 7) (x + 5) \n= (x – 7) (x + 5) px = 7, x = – 5 \nSo, zeroes of the given polynomials are \n2 + \\(\\sqrt{3}\\), 2 – \\(\\sqrt{3}\\), 7 and – 5.<\/p>\n <\/p>\n
Question 5. \nIf the polynomial x4<\/sup> – 6x3<\/sup> + 16x2<\/sup> – 25x + 10 is divided by another polynomial x2<\/sup> – 2x + k, the remainder comes out to be x + a, find k and a. \nSolution: \nIf p(x) and g(x) are any two polynomials of the form x\u00b2 – 2x and x\u00b2 + bx + c. We know by formula. \nDividend = Divisor x Quotient + Remainder \nx4<\/sup> – 6x\u00b2 + 16x3<\/sup> – 25x + 10 = (x\u00b2 – 2x + k) (x\u00b2 + bx + c) + (x + a) \nx4<\/sup> – 6x3<\/sup> + 16x\u00b2 – 25x + 10 = x4<\/sup> + bx3<\/sup> + x\u00b2c – 2x3<\/sup> – 2bx\u00b2 – 2cx + kx\u00b2 + kbx + kc + x + a \nx4<\/sup> – 6x3<\/sup> + 16x\u00b2 – 25x + 10 = x4<\/sup> + (b – 2)x3<\/sup> + (c – 2b + k) x\u00b2 + (- 2x + kb + 1)x + kc + a \nNow comparing the co-efficients of both sides b – 2 = – 6 … (1) \n[Comparing the co-efficient of x3<\/sup>] \nc – 2b + k = 16 … (2) \n[Comparing the co-efficient of x\u00b2] \n– 2c + kb + 1 = – 25 … (3) \n[Comparing the co-efficient of x] \nkc + a = 10 … (4) \n[Comparing the constant term] \nFrom (1), b = – 4 \nNow, putting the vlaue of b in equation (2) and (3) we get, \n\u21d2 c – 2(- 4) + k = 16 \nor c + k = 8 … (5) \nand -2c – 4k + 1 = – 25 \nor – 2c – 4k = – 26 \nor – c – 2k = – 13 … (6) \nAdding equation (5) and (6), we get, c + k = 8 \n– c – 2k = – 13 – k = – 5 \nSo, k = 5, \nFrom (5), c = 8 – 5 \u21d2 c = 3, \nNow put the value of k and c in equation (4), we get, \nkc + a = 10 \n5(3) + a = 10 \na = – 5 \n\u2234 Value of k and a is 5 and – 5 respectively.<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4 Question 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between …<\/p>\n
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4<\/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 2.4 - 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