# How do you find the zeros of the polynomial function with equation  f(x) = 2(x-5)(x+4)^2?

Sep 12, 2016

$x = 5$
$x = - 4$

#### Explanation:

set the equation equal to 0 and solve for each part containing an $x$.
$2 \left(x - 5\right) {\left(x + 4\right)}^{2} = 0$

$\left(x - 5\right) = 0$
${\left(x + 4\right)}^{2} = 0$ or $\left(x + 4\right) \left(x + 4\right) = 0$ so $x + 4 = 0$

this is a valid approach because its multiplicative and a $0$ in any part of the equation will result in $0$.

$x = 5$
$x = - 4$

we then plug this into the equation to verify that it works and retain the ones that do. In this equation they both are valid.