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karansingh
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If the remainder on division of x3 + 2×2 + kx +3 by x – 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x3 + 2×2 + kx – 18.

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This question is very tricky.

its an important 5 mark question for boards exam

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  1. Given, p(x) = x³ + 2x² + kx +3

    g(x) = x – 3

    Remainder, r(x=3) = 21.

    We have to find the quotient and the value of k and the zeros of the cubic polynomial.

    The division algorithm states that given any polynomial p(x) and any non-zero

    polynomial g(x), there are polynomials q(x) and r(x) such that

    p(x) = g(x) q(x) + r(x)

    To find the value of k, put x = 3 in p(x),

    p(3) = (3)³ + 2(3)² + k(3) +3 = 21

    27 + 18 + 3k + 3 = 21

    48 + 3k = 21

    3k = -27

    k = -27/3

    k = -9

    Therefore, the cubic polynomial is x³ + 2x² – 9x + 3.

    Using long division to find the quotient,

    The quotient is x² + 5x + 6.

    On factoring,

    x² + 5x + 6 = 0

    x² +3x + 2x + 6 = 0

    x(x + 3) + 2(x + 3) = 0

    (x + 2)(x + 3) = 0

    Now, x + 2 = 0

    x = -2

    Also, x + 3 = 0

    x = -3

    Therefore, the zeros of the polynomial are 3, -2 and -3

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