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a(t)=(t−k)(t−3)(t−6)(t+3) is a polynomial function of tt, where kk is a constant. Given that a(2)=0a(2)=0, what is the absolute value of the product of the zeros of aa? Answer:

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Since
a(2)=0, we know that
t-2 must be a factor of
a(t), so
k=2. Then the zeros of
a(t) are
t=2,3,6,-3, and their product is -108, whose absolute value is 108.

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