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18 February, 14:30

A (t) = (t-k) (t-3) (t-6) (t+3) Given that a (2) = 0, what is the absolute value of the product of the zeros of a?

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  1. 18 February, 14:43
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    So using a (2) = 0 we can first solve for k by substituting t for 2

    0 = (2-k) (2-3) (2-6) (2+3)

    0 = (2-k) (-1) (-4) (5)

    0 = (2-k) 20

    0 = 40 - 20k

    -40 = - 20k

    k = 2

    The next step would be to find all the 0s of a.

    0 = (t-2) (t-3) (t-6) (t+3)

    T = 2,3,6,-3

    Then we find the product

    2x3x6x-3 = - 108

    Since the problem asks for the absolute value, the answer is positive 108
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