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?

Question

28 February, 14:24

+5

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Abby Benton · Answer 1

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

Timothy · Answer 2

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.