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23 October, 14:08

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

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  1. 23 October, 16:17
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    Zeros:

    t-k=0, t=k

    t-3=0, t=3

    t-6=0, t=6

    t+3=0, t=-3

    abs value of the product = abs value of (t*3*6*-3) = 54t
  2. 23 October, 17:18
    0
    If a (2) = 0, then k=2. The product of the zeros is

    (2) * (3) * (6) * (-3) = - 108

    The absolute value of the product of zeros of a (t) is 108.
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