# How do you find all the zeros of F (x) = 5x^3 (x+ 3)^ 4 (x-7) with all its multiplicities?

Aug 20, 2017

The zeros are:

$x = 0 \text{ }$ with multiplicity $3$

$x = - 3 \text{ }$ with multiplicity $4$

$x = 7 \text{ }$ with multiplicity $1$

#### Explanation:

• Each linear factor corresponds to a zero. That is: $a$ is a zero if and only if $\left(x - a\right)$ is a factor.

• The multiplicity of each factor is the multiplicity of the corresponding zero.

We can rewrite the given $F \left(x\right)$ slightly to bring it into a form with factors with the form $\left(x - a\right)$ as follows:

$5 {x}^{3} {\left(x + 3\right)}^{4} \left(x - 7\right) = 5 {\left(x - 0\right)}^{3} {\left(x - \left(- 3\right)\right)}^{4} \left(x - 7\right)$

So the zeros are:

$x = 0 \text{ }$ with multiplicity $3$

$x = - 3 \text{ }$ with multiplicity $4$

$x = 7 \text{ }$ with multiplicity $1$