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Polynomials Test - 4

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Polynomials Test - 4
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  • Question 1
    1 / -0

    The polynomials 2x3 + ax2 + 3x – 5 and x3 + x2 – 4x – a leave the same remainder when divided by x + 1. Find the value of a out of the following options.

    Solution

    2 (-1)3 +a( -1)2+ 3(-1) – 5 = (-1)3 + (-1)2 –4(-1) – a

    -2 + a - 3 – 5 = -1 + 1 + 4 – a
    2a = 14
    a = 7

     

  • Question 2
    1 / -0

    If p(x) = 4x3 - 3x2 + 2x + 1, q(x) = x3 - x2 + x + 1 and r(x) = x2 - 2x + 1, then the value of p(x) + 2q(x) - r(x) is

    Solution

    p(x) + 2q(x) - r(x)

    = 4x3 - 3x2 + 2x + 1 + 2x3 - 2x2 + 2x + 2 - x2 + 2x - 1

    = 6x3 - 6x2 + 6x + 2

     

  • Question 3
    1 / -0

    If p + q + r = 0, then p3 + q3 + r3 equals

    Solution

    Given: p + q + r = 0
    p + q = – r ....(1)
    Cubing both sides, we get

    (p + q)3 = (–r)3

    p3 + q3 + 3pq(p + q) = – r3

    Put (p + q) = -r:
    From equation (1), we get

    p3 + q3 + r3 = 3pqr

     

  • Question 4
    1 / -0

    Factorise (3 - 4y - 7y2)2 - (4y + 1)2 and choose the correct option:

    Solution

    (3 - 4y -7y2)2 - (4y + 1)2

    (4 - 7y2) (2 - 8y - 7y2)

     

  • Question 5
    1 / -0

    If (x + 1) and (x –1) are the factors of px3 –2x + q, then the values of p and q are

    Solution

    Since (x + 1) and (x -1) are factors of px3 –2x + q

    Then p ( –1)3 – 2(1) + q = 0

    – p + q = – 2 ………(1)

    p(1)3 – 2(1) + q = 0

    p + q = 2……… (2)

    By solving we get p = 2 , q = 0

     

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