# How do you solve (x+1)^2(x+2)=0?

Dec 16, 2016

There are two solutions: ${x}_{1} = - 2 \text{ and } {x}_{2} = - 1$.

#### Explanation:

${\left(x + 1\right)}^{2} \left(x + 2\right) = 0$
Therefore,
${\left(x + 1\right)}^{2} = 0 \text{ or } x + 2 = 0$

So our first value of x is:
${x}_{1} + 2 = 0 \to {x}_{1} = - 2$

The two other values can be solved for by rewriting.
${\left(x + 1\right)}^{2} = 0 \to \left(x + 1\right) \left(x + 1\right) = 0$
Because this is a perfect square there is only one solution.
So, ${x}_{2} + 1 = 0 \to {x}_{2} = - 1$.

Our solutions are, ${x}_{1} = - 2 \text{ and } {x}_{2} = - 1$.