# How do you find all the zeros of F(x) = 8(x-6)(x+6)^5 with all its multiplicities?

Jan 23, 2018

We have two zeros $6$ with multilicity of $1$ and $- 6$ with multiplicity of $5$.

#### Explanation:

Zeros of $f \left(x\right) = k {\left(x - \alpha\right)}^{a} {\left(x - \beta\right)}^{b} {\left(x - \gamma\right)}^{c}$

are $\alpha$ with multiplicity of $a$, $\beta$ with multiplicity of $b$ and $\gamma$ with multipicity of $c$.

Hence in $f \left(x\right) = 8 \left(x - 6\right) {\left(x + 6\right)}^{5}$

we have two zeros $6$ with multilicity of $1$ and $- 6$ with multiplicity of $5$.