{"id":11244,"date":"2020-10-05T16:38:49","date_gmt":"2020-10-05T11:08:49","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=11244"},"modified":"2022-03-02T11:21:16","modified_gmt":"2022-03-02T05:51:16","slug":"polynomials-class-10-extra-questions","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/polynomials-class-10-extra-questions\/","title":{"rendered":"Polynomials Class 10 Extra Questions Maths Chapter 2 with Solutions Answers"},"content":{"rendered":"

Here you will find Polynomials Class 10 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 10 Maths<\/a> are solved by experts and will guide students in the right direction.<\/p>\n

Extra Questions for Class 10 Maths Polynomials with Answers Solutions<\/h2>\n

Extra Questions for Class 10 Maths Chapter 2 Polynomials with Solutions Answers<\/strong><\/p>\n

Polynomials Class 10 Extra Questions Objective Type<\/h3>\n

Question 1.
\nThe number of polynomials having zeroes -2 and 5 is:
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) More than 3
\nAnswer:
\n(d) More than 3<\/p>\n

Question 2.
\nIf 1 is zero of the polynomial p(x) = ax2<\/sup> – 3(a – 1)x – 1, then the value of ‘a’ is:
\n(a) 1
\n(b) -1
\n(c) 2
\n(d) – 2
\nAnswer:
\n(a) 1<\/p>\n

\"Polynomials<\/p>\n

Question 3.
\nIf a, \u00df are zeroes of x2<\/sup> – 6x + k. What is the value of k if 3a + 2B = 20.
\n(a) – 16
\n(b) 8
\n(c) – 2
\n(d) -8
\nAnswer:
\n(a) – 16<\/p>\n

Question 4.
\nIf one zero of 2x2<\/sup> – 3x + k is reciprocal to the other, then the value of k is:
\n(a) 2
\n(b) \\(\\frac {-2}{3}\\)
\n(c) \\(\\frac {-3}{2}\\)
\n(d) -3
\nAnswer:
\n(a) 2<\/p>\n

Question 5.
\nIf p(x) = x2<\/sup> + 6x + 9 and q(x) = x + 3 then remainder will be when p(x) is divided by q(x):
\n(a) – 1
\n(b) 0
\n(c) 11
\n(d) 2
\nAnswer:
\n(b) 0<\/p>\n

Question 6.
\nDividing (x2<\/sup> + 1) by (x + 1), the remainder will be:
\n(a) -1
\n(b) 11
\n(c) 0
\n(d) – 2
\nAnswer:
\n(c) 0<\/p>\n

Question 7.
\nDividing x3<\/sup> + 3x + 3 by (x + 2), the remainder will be:
\n(a) -2
\n(b) -1
\n(c) 0
\n(d) 1
\nAnswer:
\n(d) 1<\/p>\n

\"Polynomials<\/p>\n

Question 8.
\nThe zero’s of the polynomial (x2<\/sup> – 2x – 3) will be:
\n(a) – 3, 1
\n(b) -3, -1
\n(c) 3, -1
\n(d) 3, 1
\nAnswer:
\n(c) 3, -1
\nSolution:
\nx2<\/sup> – 2x – 3 = x2<\/sup> – 3x + x – 3
\n= x(x – 3) + 1 (x – 3 ) = (x – 3) (x + 1)
\n\u2234 Zeros of the polynomial are 3, -1
\nHence, choice (c) is correct.<\/p>\n

Question 9.
\nDividing x3<\/sup> – 3x2<\/sup> – x + 3 by x – 4x + 3 the remainder will be:
\n(a) – 3
\n(b) 3
\n(c) 1
\n(d) 0
\nAnswer:
\n(d) 0<\/p>\n

Polynomials Class 10 Extra Questions Very Short Answer Type<\/h3>\n

Question 1.
\nFind a quadratic polynomial each with the given numbers as the sum and product of the zeroes respectively.
\n(i) \\(\\frac {1}{4}\\), -1
\n(ii) \u221a2, \\(\\frac {1}{3}\\)
\n(iii) 0, \u221a5
\n(iv) 1, 1
\n(v) – \\(\\frac {1}{4}\\), \\(\\frac {1}{4}\\)
\n(vi) 4, 1
\nSolution:
\nLet the polynomial be ax2<\/sup> + bc + c and its zeroes be \u03b1 and \u03b2.
\n(i) Here, \u03b1 + \u03b2 = \\(\\frac {1}{4}\\) and \u03b1\u03b2 = -1
\nThus, the polynomial formed = x2<\/sup> – (Sum of the zeroes)x + Product of the zeroes
\n\"Polynomials
\nIf k = 4 then the polynomial is 4x2<\/sup> – 3 – 4.<\/p>\n

(ii) Here, \u03b1 + \u03b2 = \u221a2 and \u03b1\u03b2 = \\(\\frac {1}{3}\\)
\nThus, the polynomial formed = x2<\/sup> – (Sum of the zeroes) x + Product of the zeroes
\n\"Polynomials
\nIf k = 3, then, the polynomial is 3x2<\/sup> – 3\u221a2x + 1.<\/p>\n

(iii) Here, \u03b1 + \u03b2 = 0 and a\u00df = \u221a5
\nThus, the polynomial formed = x2<\/sup> – (Sum of the zeroes) x + Product of zeroes
\n= x2<\/sup> – (0)x + \u221a5
\n= x2<\/sup> + \u221a5<\/p>\n

(iv) Let the polynomial be ax2<\/sup> + bx + c and its zeroes \u03b1 and \u03b2. Then
\n\"Polynomials
\nIf a = 1 then b = -1 and c = 1
\n\u2234 One quadratic polynomial which satisfy the given conditions is x2<\/sup> – x + 1.<\/p>\n

(v) Let the polynomial be ax2<\/sup> + bx +c and its zeroes be a and B. Then,
\n\"Polynomials
\nif a = 4, then b = 1 and c = 1.
\n\u2234 One quadratic polynomial which satisfy the given conditions is 4x2<\/sup> + x + 1.<\/p>\n

(vi) Let the polynomial be ax2<\/sup> + bx + c and its zeroes be a and \u00df. then,
\n\"Polynomials
\nand \u03b1\u03b2 = 1 if a = 1 then b = -4 and c = 1
\n\u2234 One quadratic polynomial which satisfy the given conditions is x2<\/sup> – 4x + 1.<\/p>\n

\"Polynomials<\/p>\n

Question 2.
\nDivide p(y) by g(y) if p(y) = y3<\/sup> – 3y2<\/sup> – y + 3 and g(y) = y2<\/sup> – 4y + 3.
\nAnswer:
\nWe write
\ny3<\/sup> – 3y2<\/sup> – y + 3 = (y + 1) (y2<\/sup> – 4y + 3).
\nSolution:
\n\"Polynomials
\nHere, the quotient is y + 1 and the remainder is zero.<\/p>\n

Question 3.
\nExamine if x – 1 is a factor of 2x3<\/sup> – 5x + 3.
\nSolution:
\n\"Polynomials
\nHere, remainder is 0, hence, x – 1 is one factor of 2x3<\/sup> – 5x + 3.<\/p>\n

Polynomials Class 10 Extra Questions Short Answer Type<\/h3>\n

Question 1.
\nDivide the polynomial f(x) = 3x2<\/sup> – x3<\/sup> – 3x + 5 by the polynomial g(x) = x – 1 – x2<\/sup> and verify the division alogrithm.
\nSolution:
\nf(x) = – x3<\/sup> + 3x2<\/sup> – 3x + 5 and g(x) = – x2<\/sup> + x- 1
\n\"Polynomials
\n\u2234 Quotient = x – 2 and remainder = 3.
\nas dividend = Q x divisior + remainder
\n– x3<\/sup> + 3x2<\/sup> – 3x + 5 = (x – 2)(-x2<\/sup> + x – 1) +3
\n= – x3<\/sup> + x2<\/sup> – x + 2x2<\/sup> – 2x + 2 + 3
\n= – x3<\/sup> + 3x2<\/sup> – 3x + 5
\nVerify<\/p>\n

\"Polynomials<\/p>\n

Question 2.
\nDivide 3 – x + 2x2<\/sup> by (2 – x) and verify alogrithm.
\nSolution:
\nFirst we write the terms of dividend and divisor in decreasing order of their degrees and then perform the division as shown below:
\n\"Polynomials
\nClearly, degree (9) = 0 < degree (- x + 2).
\n\u2234 quotient = (- 2x – 3) and remainder = 9.
\n= (quotient x divisor) + remainder
\n= (- 2x – 3)(- x + 2) + 9
\n= 2x2<\/sup> – 4x + 3x – 6 + 9
\n= 2x2<\/sup> – x + 3
\n= dividend
\nThus, (quotient x divisor) + remainder = dividend. Hence, the division algorithm is verified.<\/p>\n

\"Polynomials<\/p>\n

Question 3.
\nOn dividing x3<\/sup> – 3x2<\/sup> + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectvely. Find g(x).
\nSolution:
\nGiven, p(x) = x2<\/sup> – 3x2<\/sup> + x + 2, q(x) = x – 2 and r(x) = – 2x + 4. By division algorithm, we know that
\nDividend = Divisor \u00d7 Quotient + Remainder
\np(x) = q(x) \u00d7 g(x) + r(x).
\nTherefore,
\nx3<\/sup> + 3x2<\/sup> + x + 2 = (x – 2) \u00d7 g(x) + (-2x + 4)
\n\u21d2 x3<\/sup> – 3x2<\/sup> + x + 2 + 2x – 4 = (x – 2) \u00d7 g(x)
\n\u21d2 \"Polynomials
\nOn dividing x3<\/sup> – 3x2<\/sup> + 3x – 2 by x – 2, we get
\n\"Polynomials
\nHence, g(x) = x2<\/sup> – x + 1.<\/p>\n

\"Polynomials<\/p>\n

Question 4.
\nDivide 6x5<\/sup> + 5x4<\/sup> + 11x3<\/sup> – 5x2<\/sup> + 2x + 7 by 3x2<\/sup> – 2x + 4.
\nSolution:
\n\"Polynomials<\/p>\n

Question 5.
\nIf – 3 is one of the zeros of the quadratic polynomial (k – 1) x2<\/sup> + kx + 1. Find the value of other zeros.
\nSolution:
\nGiven that – 3 is one zeros of the polynomial
\n(k – 1) x2<\/sup> + kx + 1
\n\u2234 (k – 1)(-3)2<\/sup> + k (-3) + 1 = 0 .
\n\u21d2 9k – 9 – 3k + 1 = 0
\n\u21d2 6k = 8
\n\u21d2 Putting the value of k = \\(\\frac {4}{3}\\) in given polynomial,
\nwe get (\\(\\frac {4}{3}\\)– 1) x2<\/sup> + \\(\\frac {4}{3}\\) x + 1
\n\"Polynomials
\n\u21d2 x2<\/sup> + 4x + 3 = x2<\/sup> + 3x + x + 3
\n= x (x + 3) + 1 (x + 3)
\n\u21d2 (x + 3) (x + 1)
\n\u2234 zeros are – 3 and – 1
\nHence their zeros of the quadratic polynomial is – 1.<\/p>\n

\"Polynomials<\/p>\n

Question 6.
\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 g(x)
\n(ii) deg g(x) = deg r(x)
\n(iii) deg q(x) = 0
\nSolution:
\nLet
\nq(x) = 3x2<\/sup> + 2x + 6,
\ndegree of g(x) = 2
\np(x) = 12x2<\/sup> + 8x + 24,
\ndegree of p(x) = 2<\/p>\n

(i) Using division algorithm,
\nWe have,’ p(x) = (x) \u00d7 g(x) + r(x)
\nOn dividing 12x2<\/sup> + 8x + 24 by 3x2<\/sup> + 2x + 6.
\nwe get
\n\"Polynomials
\nSince, the remainder is zero, therefore 3x2<\/sup> + 2x + 6 is a factor of 12x2<\/sup> + 8x + 24.
\n\u2234 g(x) = 4 and r(x) = 0.<\/p>\n

(ii) p(x) = x5<\/sup> + 2x4<\/sup> + 3x3<\/sup> + 5x2<\/sup> + 2
\nq(x) = x2<\/sup> + x + 1, degree of g(x) = 2
\ng(x) = x3<\/sup> + x2<\/sup> + x + 1
\nr(x) = 2x2<\/sup> – 2x + 1, degree of r(x) = 2
\nHere, deg q(x) = deg r(x)
\nOn dividing x5<\/sup> + 2x4<\/sup> + 3x3<\/sup> + 5x2<\/sup> + 2 by x2<\/sup> + x + 1, we get
\n\"Polynomials
\nHere, g(x) = x3<\/sup> + x2<\/sup> + x + 1
\nand r(x) = 2x2<\/sup> – 2x + 1<\/p>\n

(iii) Let p(x) = 2x4<\/sup> + 8x3<\/sup> + 6x2<\/sup> + 4x + 12, r(x) = 2, degree of r(x) = 0
\ng(x) = x4<\/sup> + 4x3<\/sup> + 3x2<\/sup> + 2x + 1
\n\u21d2 9(x) = 10
\nHere deg r(x) = 0
\nOn dividing 2x4<\/sup> + 8x3<\/sup> + 6x2<\/sup> + 4x + 12 by 2,
\nwe get
\n\"Polynomials
\n\"Polynomials<\/p>\n

\"Polynomials<\/p>\n

Question 7.
\nFind the zero’s of quadratic polynomial f(x) = 3x2<\/sup> – 3 – 4. Verify the relationship between the zeros and its coefficients.
\nSolution:
\n\u2234 f(x) = 3x2<\/sup> – x – 4
\n= 3x2<\/sup> – 4x + 3x – 4
\n= x(3x – 4) + 1 (3x – 4)
\n= (3x – 4) (x + 1)
\n\u2234 zero’s are \\(\\frac {4}{3}\\) and -1.
\nsum of the zeros =
\n\"Polynomials<\/p>\n

Question 8.
\nSolve the pair of linear simultaneous equations
\n2x – y = 1 and x + 2y = 13
\nBy drawing their graphs. Find the coordinates of vertices of a triangle formed by these lines and y-axis.
\nSolution:
\n2x – y = 1 \u21d2 x + 2y = 13
\n2x = 1 + y \u21d2 x = 13 – 2y
\n\"Polynomials
\nNow plot the points on the graph paper.
\n\"Polynomials
\nThe coordinate of required \u2206 are
\nA(3, 5), B (0, \\(\\frac {13}{2}\\)) and C(0, -1) respectively
\nSolution is x = 3, y = 5<\/p>\n

\"Polynomials<\/p>\n

Question 9.
\nFind all the zeros of polynomial 2x4<\/sup> – 3x2<\/sup> – 3x2<\/sup> + 6x – 2. If two of its zeros are \u221a2 and \u221a2.
\nSolution.
\nquad. polynomial form by the given zeros is
\n(X – \u221a2) (x + \u221a2)
\n\u21d2 x2<\/sup> – 2
\nNow divided the given polynomial by x2<\/sup> – 2
\n\"Polynomials
\n\u2234 x4<\/sup> – 3x3<\/sup> – 3x2<\/sup> + 6x – 2
\n= (x2<\/sup> – 2) (2x2<\/sup> – 3x + 1)
\n= (x2<\/sup> – 2) (2x2<\/sup> – 2x – x + 1)
\n= (x2<\/sup> – 2) [2x(x -1) – 1(x-1)]
\n= (x \u221a2 )(x + \u221a2)(x – 1) (2x – 1)
\nHence all zeros are
\n1\/2, 1, \u221a2 and – \u221a2<\/p>\n

\"Polynomials<\/p>\n","protected":false},"excerpt":{"rendered":"

Here you will find Polynomials Class 10 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Polynomials with Answers Solutions Extra Questions for Class 10 Maths Chapter 2 Polynomials with Solutions Answers …<\/p>\n

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