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NCERT Solutions 10th Maths Chapter 1 Polynomials Exercise 1.3
If you are interested in reading NCERT Solutions for Chapter 2 Class 10 Polynomials then you are at the right frame of mind. You will get all subjects correct and to the point CBSE NCERT Solutions. manabadi.coin provides these strategies that will aid you a great deal in scoring good marks in the exams.
Exercise 2.3 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: Answer (i) p(x) = x3 ñ 3x2 + 5x ñ 3, g(x) = x2 ñ 2 Quotient = x-3 and remainder 7x - 9 (ii) p(x) = x4 ñ 3x2 + 4x + 5, g(x) = x2 + 1 ñ x Quotient = x2 + x - 3 and remainder 8 (iii) p(x) = x4 ñ 5x + 6, g(x) = 2 ñ x2 Quotient = -x2 -2 and remainder -5x +10 2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: Answer (i) t2 ñ 3, 2t4 + 3t3 ñ 2t2 ñ 9t ñ 12 t2 ñ 3 exactly divides 2t4 + 3t3 ñ 2t2 ñ 9t ñ 12 leaving no remainder. Hence, it is a factor of 2t4 + 3t3 ñ 2t2 ñ 9t ñ 12. (ii) x2 + 3x + 1, 3x4 + 5x3 ñ 7x2 + 2x + 2 x2 + 3x + 1 exactly divides 3x4 + 5x3 ñ 7x2 + 2x + 2 leaving no remainder. Hence, it is factor of 3x4 + 5x3 ñ 7x2 + 2x + 2. (iii) x3 ñ 3x + 1, x5 ñ 4x3 + x2 + 3x + 1 x3 ñ 3x + 1 didn't divides exactly x5 ñ 4x3 + x2 + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of x5 ñ 4x3 + x2 + 3x + 1. 3. Obtain all other zeroes of 3x4 + 6x3 ñ 2x2 ñ 10x ñ 5, if two of its zeroes are v(5/3) and - v(5/3). Answer p(x) = 3x4 + 6x3 ñ 2x2 ñ 10x ñ 5 Since the two zeroes are v(5/3) and - v(5/3). We factorize x2 + 2x + 1 = (x + 1)2 Therefore, its zero is given by x + 1 = 0 x = -1 As it has the term (x + 1)2 , therefore, there will be 2 zeroes at x = - 1. Hence, the zeroes of the given polynomial are v(5/3) and - v(5/3), - 1 and - 1. 4. On dividing x3 - 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and -2x + 4, respectively. Find g(x). Answer Here in the given question, Dividend = x3 - 3x2 + x + 2 Quotient = x - 2 Remainder = -2x + 4 Divisor = g(x) We know that, Dividend = Quotient ◊ Divisor + Remainder ? x3 - 3x2 + x + 2 = (x - 2) ◊ g(x) + (-2x + 4)? x3 - 3x2 + x + 2 - (-2x + 4) = (x - 2) ◊ g(x) ? x3 - 3x2 + 3x - 2 = (x - 2) ◊ g(x) ? g(x) = (x3 - 3x2 + 3x - 2)/(x - 2) ? g(x) = (x2 - x + 1) 5.Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0 Answer (i) Let us assume the division of 6x2 + 2x + 2 by 2 Here, p(x) = 6x2 + 2x + 2 g(x) = 2 q(x) = 3x2 + x + 1 r(x) = 0 Degree of p(x) and q(x) is same i.e. 2. Checking for division algorithm, p(x) = g(x) ◊ q(x) + r(x) Or, 6x2 + 2x + 2 = 2x (3x2 + x + 1) Hence, division algorithm is satisfied. (ii) Let us assume the division of x3+ x by x2, Here, p(x) = x3 + x g(x) = x2 q(x) = x and r(x) = x Clearly, the degree of q(x) and r(x) is the same i.e., 1. Checking for division algorithm, p(x) = g(x) ◊ q(x) + r(x) x3 + x = (x2 ) ◊ x + x x3 + x = x3 + x Thus, the division algorithm is satisfied. (iii) Let us assume the division of x3+ 1 by x2. Here, p(x) = x3 + 1 g(x) = x2 q(x) = x and r(x) = 1 Clearly, the degree of r(x) is 0. Checking for division algorithm, p(x) = g(x) ◊ q(x) + r(x) x3 + 1 = (x2 ) ◊ x + 1 x3 + 1 = x3 + 1 Thus, the division algorithm is satisfied.Post your comments
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