# How do you solve (x+4)^3(x+2)^2=0?

Jun 14, 2017

$x = - 4 , - 2$

#### Explanation:

We have: ${\left(x + 4\right)}^{3} {\left(x + 2\right)}^{2} = 0$

Using the null factor law:

$R i g h t a r r o w {\left(x + 4\right)}^{3} = 0$

$R i g h t a r r o w x + 4 = 0$

$\therefore x = - 4$

$\mathmr{and}$

$R i g h t a r r o w {\left(x + 2\right)}^{2} = 0$

$R i g h t a r r o w x + 2 = 0$

$\therefore x = - 2$

Therefore, the solutions to the equation are $x = - 4$ and $x = - 2$.