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

Aug 11, 2016

x=0, &, x=-4, each with single multiplicity; $x = 2 ,$ with multiplicity$= 3$

#### Explanation:

To find the zeroes of a poly. $f \left(x\right)$, we need to solve the eqn. $f \left(x\right) = 0$

In our case, $f \left(x\right) = 0 \Rightarrow x \left(x + 4\right) {\left(x - 2\right)}^{3} = 0$

$\Rightarrow x = 0 , \mathmr{and} , x + 4 = 0 , \mathmr{and} , x - 2 = 0$

rArr x=0, &, x=-4, each with single multiplicity; $x = 2 ,$ with multiplicity$= 3$