Chapter 2 Polynomials class9 Practice Questions Paper Important Identities : - 1. ( x + y )2 = x2 + 2xy +y2 2. ( x – y)2 = x2 – 2xy + y2 3. (x + y)(x – y) = x2 – y2 4. (x + a)(x + b) = x2 +(a + b)x + ab 5. (x + y)3 = x3 + 3x2y + 3xy2 + y3 = x3 + y3 +3xy(x +y) 6. (x - y)3 = x3 - 3x2y + 3xy2 - y3 = x3+ y3 -3xy(x -y) 7. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx 8. x3 + y3 = (x + y)(x2 – xy + y2) x3 - y3 = (x - y)(x2 + xy + y2) 9. x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) 10. If x + y + z = 0 , then x3 + y3 + z3 = 3xyz 1. Classify the following as monomials, binomials and trinomials : (a)x3 b) 2y2 – 4y + 3 c) t2 – 4 d) √2 e) x3 + 4x2 + 5x f) u7 + u2 – 4. 2. Write the coefficients of x2 in each of the following : a) 3x^2 – 4y b) x + x^2 + 7y c) 3x + 4y – 5z d) x^2 + 2xy + 3y^2 3. Write the degree each of the following : a) 5x^3 + 4x^2 + 7x b) 4 – y2 c) 5t – 3 4. Classify the following as linear, quadratic and cubic polynomials : a) x^2 + x b) x – x^3 c) y + y^2 + 4 d) 1 + x e) 3t f) r^2 5. Find the value of the polynomial 5x – 4x^2 + 3 at : a) x = 0 b) x = - 1 c) x = 2 6. Find the value of each of the following polynomials at indicated value of variables : a) p(x) = 5x^2 – 3x + 7 at x = 1 b) p(y) = 3y^3 – 4y + 4 at y = 2 c) p(t) = 4t^4 + 5y^3 – t^2 + 6 at t = a 7. Check whether – 2 and 2 are zeroes of the polynomial x + 2 . 8. Find the zero of the polynomial p(x) = 2x + 1. 9. Verify whether 2 and - 2 are zero of the polynomial x^2 – 4 . 10. Verify whether 2 and 0 are zero of the polynomial x^2 – 2x 11. Find the value of the following : a) (3x2 – 3x + 1)(x – 1) when x = 3 b) (3x2 – 1)(4x3 – 4x – 3) when x = - 1. 12. Evaluate the following for given values of the variables : a) x4 – x3 + x2 – x + 1 for x = 2 b) x3 + x2 + x + 1 for x = - 1. 13. Find the remainder and quotient in each of the following a) Divide x4 – 1 by x – 1 . b) Divide x3 – 3x2 + 5x – 8 by x – 2. 14. Find the remainder when 4x3 – 3x2 + 2x – 4 is divided by : a) x – 1 b) x – 2 c) x + 1 d) x – 4 e) x + 2 15. Using remainder theorem, find the remainder : a) Divide x6 – 1 by x – 1 b) Divide x3 + 1 by x + 1 . 16. Find the remainder when x4 + x3 – 2x2 + x + 1 is divided by x + 1. 17. Find the remainder when the polynomial p(x) = x3 + 2x2 – 2x + 1 is divided by x + 3 18. Find the remainder when the polynomial p(x) = x2 +4 x + 2 is divided by x + 2. 19. Find the remainder when 3x4 - x3 + 3x2 - 4 x + 1 is divided by x – 3. 20. If x – 2 is a factor of each of the following polynomials, then find the value of a in each case : a) x2 – 3x + 5a b) x3 – 2ax2 + ax – 1 c) x5 – 3x4 – ax3 + 3ax2 + 2ax + 4. 21. Factorise : 6x2 + 17x + 5 22. Factorise : x3 – 23x2 + 142x – 120 . 23. Using a suitable identity, find the following products : a) (x + 5)(x – 3) b) (4x + 3)(4x + 5) c) (x + y)(x + y) e) (3x + 4)(3x + 7) f) (5a + 3)(5a + 2) 24. Expand using suitable formula : a) (2a + 3)2 b) (3a – 5)2 25. Factorise the following : a) x2 + 6x + 9 b) 24x2 – 41x + 12 c) x2 – x – 6 d) 16x2 + 8x + 1 e) 9x2 – 16y2 f) 4x3 – 4x g) (x + 1)2 – (x –1)2 h) 9x2 + 6x + 1 – 25y2 i) 25x2 – 10x + 1 – 36y2 j) x3 + x – 3x2 – 3 k) x2 + y – xy – x l) 3ax – 6ay – 8by + 4bx m) xy – ab + bx – ay n) 1 – x2 – y2 – 2xy o) 8 – 4a – 2a3 + a4 p) a2 + b2 + 2ab + 2bc + 2ca q) x3 + 64 r) 25x2 – 10x + 1 s) x2 – 11x – 42 t) 12x2 – 10x + 2 v) a4 – a w) x3 – 125 x)27x3y3 – 8z3 y) 8x3 – (2x – y)3 z) (a + b)3 – (a – b)3 26. factorise the following : i) 4(x – y)2 – 12(x – y)(x + y) + 9(x + y)2 ii) 3(x + y)2 – 5(x + y ) + 2 iii) 12(x2 + 7x)2 – 8(x2 + 7x)(2x – 1) – 15(2x – 1)2 iv) x2 – 5x + 6 v)4x2 + 9y2 + z2 + 12xy + 4xz + 6yz vi) x3 – x2y + xy2 – y3 vii) a3 +a2b +ab2 + b3 27. Solve using appropriate formula : i) (2a + 3)(2a – 3) ii) (105)2 iii)(49)2 iv) (536)2 – (136)2 v) if 4x = 72 – 32, then find the value of x. vi) 998 × 1002 28. Simplify : (a + b)3 + (a – b)3 + 6a(a2 – b2). 29. Show that if 2(a2 + b2) = (a + b)2, then a = b. 30. Expand each of the following : - i) (x + 2y)3 ii) (2x – 3y)3 iii) (x2 + 2y)3 31. Evaluate the following using suitable identities : i) (98)3 ii) (101)3 iii) (999)3 32. Show that if (a + b) is not zero, then the equation : a(x – a) = 2ab – b(x – b) has a solution x = a + b. 33. Foctorise each of the following : i) a4 – b4 ii) a4 – 16b4 iii) a2 – (b – c)2 iv) x2 + 7xy + 12y2 v) x2 + 2ax – b2 – 2ab vi) (x2 + x)2 + 4(x2 + x) – 12 vii) 5x2 + 16x + 3. From more study Material JSunil Tutorial |
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