Polynomials - Test Papers

 CBSE Test Paper 01

CH-2 Polynomials


  1. If p(x) = x + 3, then p(x) + p(-x) is equal to
    1. 2x
    2. 3
    3. 0
    4. 6
  2. If x+y+z = 0, thenx3+y3+z3 is 
    1. 3xyz
    2. xyz
    3. 2xyz
    4. 0
  3. The degree of constant function is
    1. 0
    2. 3
    3. 1
    4. 2
  4. (x + 1) is a factor of the polynomial
    1. x3+x2x+1
    2. x3+x2+x+1
    3. x4+3x3+3x2+x+1
    4. x4+x3+x2+1
  5. The coefficient of ‘x’ in the expansion of (x+3)3 is
    1. 1
    2. 27
    3. 9
    4. 18
  6. Fill in the blanks: A polynomial containing one non-zero term is called a ________.
  7. Fill in the blanks: The highest power of the variable in a polynomial is called the ________ of the polynomial.
  8. Write (-3x + y + z)2 in the expanded form:
  9. Evaluate the following by using identities: 0.54 × 0.54 - 0.46 × 0.46
  10. Evaluate: 185 × 185 - 15 × 15
  11. Evaluate 105 × 108 without multiplying directly.
  12. Simplify (x + y)3 - (x - y)3 - 6y(x + y) (x - y).
  13. Factorize : (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3
  14. If both x - 2 and x12 are factors of px2 + 5x + r, show that p = r.
  15. If the polynomials az3 + 4z2 + 3z - 4 and z3 - 4z + a leave the same remainder when divided by z - 3, find the value of a.

CBSE Test Paper 01
CH-2 Polynomials


Solution

  1. (d) 6
    Explanation:
    p(x) = x + 3
    And p(-x) = -x + 3
    Then, p(x) + p(-x)
    = x + 3 - x + 3
    = 6
  2. (a) 3xyz
    Explanation: x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzzx) =>x3+y3+z33xyz=(0)(x2+y2+z2xyyzzx) => x3+y3+z33xyz=0 => x3+y3+z3=3xyz If x+y+z = 0, thenx3+y3+z3 is  3xyz
  3. (a) 0
    Explanation: The degree of any constant term 5 (say)
    We can write 5 as 5 x 1 = 5xo [Since ao= 1]
    Therefore the degree of any constant term is 0
  4. (b) x3+x2+x+1
    Explanation: x3+x2+x+1 = x3(x+1)+1(x+1) = (x3+1)(x+1)
  5. (b) 27
    Explanation:(x+3)3   = x3+(3)3+3×x×3(x+3) = x3+27+9x2+27x = x3+9x2+27x+27 Therefore, the coefficient of x, in the expansion of (x+3)3 is 27.
  6. monomial
  7. degree
  8. We have,
    (-3x + y + z)2
    Using identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
    = (-3x)2 + y2 + z2 + 2 × (-3x) × y + 2 × y × z + 2 × z × (-3x)
    = 9x2 + y2 + z2 - 6xy + 2yz - 6zx
  9. We have,
    0.54 × 0.54 - 0.46 × 0.46
    = (0.54)2 - (0.46)2 = (0.54 + 0.46)(0.54 - 0.46) = 1 × 0.08 = 0.08
  10. 185 ×185 - 15 × 15
    (185)2(15)2 [Using a2b2=(ab)(a+b)]
    (185+15)(18515)
    200 × 170 = 34000
  11. 105× 108=(100+5)(100+8)
    Using identity (x+a)(x+b)=x2+(a+b)x+ab
    We get, 105 × 108 = 1002 + (5 + 8)100 + 5 × 8
    =10000+1300+40=11340
  12. (x + y)- (x - y)3 - 6y(x + y)(x - y)
    = (x + y)3 - (x - y)3 - 3.2y(x + y)(x - y)
    = (x + y)3 - (x - y)3 - 3 (x + y - x + y)(x + y)(x - y)
    = [x + y - x + y)]3 [ a- b3 - 3ab(a - b) = (a - b)3]
    =(2y)3 = 8y3
  13. Let x = a2 - b2, y = b2 - c2 and z = c2 - a2. Then,
    x + y + z = a2 - b2 + b2 - c2 + c2 - a2 = 0
     x3 + y3 + z3 = 3xyz
     (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a2 - b2)(b2 - c2)(c2 - a2)
     (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a + b)(a - b)(b + c)(b - c)(c + a)(c - a)
     (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a + b)(b + c)(c + a)(a - b)(b - c)(c - a)
  14. Let f(x) = px2 + 5x + r be the given polynomial. Since x - 2 and x12are factors of f(x).
     f(2) = 0 and f(12) = 0
     p × 2+ 5 × 2 + r = 0 and p(12)+ 5 ×12 + r = 0
     4p + 10 + r = 0 and p4+52 + r = 0
     4p + r = -10 and p+4r+104=0
     4p + r = -10 and p + 4r + 10 = 0
     4p + r = -10 and p + 4r = -10
     4p + r = p + 4r [RHS of the two equations are equal]
     3p = 3r  p = r
  15. Let p(z) = az3 + 4z2 + 3z - 4
    And q(z) = z3 - 4z + a
    As these two polynomials leave the same remainder, when divided by z - 3, then p(3) = q(3).
     p(3) = a(3)3 + 4(3)2 + 3(3) - 4
    = 27a + 36 + 9 - 4
    Or p(3) = 27a + 41
    And q(3) = (3)3 - 4(3) + a
    = 27 - 12 + a = 15 + a
    Now, p(3) = q(3)
     27a + 41 = 15 + a
     26a = -26; a = -1
    Hence, the required value of a = -1.