POLYNOMIALS CLASS 9TH TO 12TH CBSE BOARD, Exams of Mathematics

Download POLYNOMIALS CLASS 9TH TO 12TH CBSE BOARD and more Exams Mathematics in PDF only on Docsity! CBSE Test Paper 02 CH-2 Polynomials 1. The factors of is a. b. c. d. 2. A polynomial of degree ____ is called a linear polynomial. a. 1 b. 2 c. 3 d. 0 3. Which of the following is a polynomial in one variable? a. b. c. d. 4. If p(x) = (x – 1)(x + 1), then the value of p(2) + p(1) – p(0) is a. 2 b. 4 c. 1 d. 3 5. If both and are the factors of , then a. none of these b. 2p = r c. p = r d. p = 2r 6. Fill in the blanks: The maximum number of terms in a polynomial of degree 10 is ________. 7. Fill in the blanks: is a polynomial of degree ________. 8. Find p(0), p(1) and p(2) of the polynomial: 9. Factorize: x4 + x2 + 1 10. If x2 -1 is a factor of ax3+ bx2+ cx + d, show that a + c =0. 11. Check whether is a factor of . 12. Factorise: 84 - 2r - 2r2 13. Factorize the polynomial: 14. Factorise: 4x2 + 20x + 25 15. The polynomial 3x3+ ax2 + 3x + 5 and 4x3 + x2 - 2x + a leave remainder when divided by (x - 2) respectively. If R1 - R2 = 9, find the value of a. 9. We have, x4 + x2 + 1 = (x4 + 2x2 + 1) - x2 = (x2 + 1)2 - x2 = (x2 + 1 - x)(x2 + 1 + x) = (x2 - x + 1)(x2 + x + 1) 10. Since is a factor of a + b + c + d = -a + b - c + d = 0 a + c = 0 11. We know that if the polynomial is a factor of , then on dividing the polynomial by , we must get the remainder as 0. We need to find the zero of the polynomial 7 + 3x While applying the remainder theorem, we need to put the zero of the polynomial in the polynomial ,to get We conclude that on dividing the polynomial by 7 + 3x, we will get the remainder as which is not 0. Therefore, we conclude that 7 + 3x is not a factor of 12. In order to factorise 84 - 2r - 2r2, we have to find two numbers p and q such that p + q = -2 and pq = - 168. Clearly, (-14) + 12 = -2 and (-14) × 12 = -168. So, we write the middle term - 2r as (-14r) + 12r. 84 - 2r - 2r2 = -2r2 - 2r + 84 = -2r2 - 14r + 12r + 84 = -2r(r + 7) + 12(r + 7) = (r + 7)(-2r + 12) = -2(r + 7)(r - 6) = -2(r - 6)(r + 7) 13. The expression can be written as Using identity Therefore, after factorizing the expression 14. We have, 4x2 + 20x + 25 = (2x)2 + 2(2x)(5) + (5)2 = (2x + 5)2 [ a2 + 2ab + b2 = (a + b)2] = (2x + 5)(2x + 5) 15. Let f (x) = 3x3 + ax2 + 3x + 5 and g(x) = 4x3+ x2 - 2x + a Here, the zero of (x - 2) is x = 2 [ x - 2 = 0 x = 2] Where f(x) and g(x) are divided by (x - 2), then we get the remainders R1 and R2 f(2) = 3(2)3 + a(2)2 + 3(2) + 5 = 24 + 4a + 6 + 5 = 35 + 4a = R1 and g(2) = 4(2)3 + (2)2 - 2(2) + a = 32 + 4 - 4 + a = 32 + a = R2 Also, R1 - R2 = 9 35 + 4a - (32 + a) = 9 3 + 3a = 9 3a = 6 a = 2