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