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19 October, 23:37

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

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  1. 20 October, 00:15
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    Let p (x) = x^3+2x^2+kx+3

    On dividing p (x) by x-3, the remainder is 21. Therefore,

    P (3) = 21

    Substituting x=3 in p (x)

    P (3) = 3^3 + 2 * (3) ^2+k*3+3

    =27+18+3k+3

    =48+3k

    We know that, p (3) = 21.

    So, 48+3k=21

    3k=21-48

    3k=-27

    k=-27/3 = - 9

    now, p (x) = x^3+2x^2-9x-18

    -2 is a factor of p (x) on inspection. Therefore, divide p (x) by x+2 to find the

    zeroes of the polynomial.

    On dividing, we get the factors to be, (x^2-9) (x+2)

    (x^2-3^2) (x+2)

    Factorizing using the identity a^2-b^2 = (a+b) (a-b) we get,

    (x+3) (x-3) (x+2)

    Therefore, the zeroes of the polynomials are - 3,+3 and - 2.
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