# How do you find all the zeros of F(x) = -4 (x+7)^3 (x-7)^2 with all its multiplicities?

Oct 10, 2016

$F \left(x\right)$ has zeros at $\left(- 7\right)$ with multiplicity of $3$ and at (+7 with a multiplicity of $2$

#### Explanation:

$F \left(x\right) = - 4 {\left(x + 7\right)}^{3} \left(x - 7\right)$^2

A term of $\left(x + 7\right)$ implies a zero at $x = - 7$
A term of $\left(x + 7\right)$ implies a zero at (x=+7#

$F \left(x\right) = - 4 \cdot {\underbrace{\left(x + 7\right) \left(x + 7\right) \left(x + 7\right)}}_{\text{multiplicity of 3" * underbrace((x-7)(x-7))_"multiplicity of 2}}$