(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?
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Hey vaironxxrd. Your answers are spot on.
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