Thread: Polynomial Function

(a) List each Zero for ... $\displaystyle F(x) = 3(x-7)(x+3)^2$

I did the following $\displaystyle F(x) = \frac{3(x-7)(x+3)^2}{3} = \frac{0}{3}$

Now $\displaystyle (x-7)(x+3)(x+3) = 0$
The zeros are 7,-3 Multiplicity 2.

Correct?

Hey vaironxxrd.

Your answers are spot on.