# How do you solve 9x ^ { 4} + 18x ^ { 3} + 9x ^ { 2} = 0?

Oct 25, 2017

$x = - 1 , - 1 , 0 , 0$

#### Explanation:

First, begin by factoring as much as possible from each term of the equation on the left hand side:

$9 {x}^{4} + 18 {x}^{3} + 9 {x}^{2} = 0$

$\left(9 {x}^{2}\right) \left({x}^{2} + 2 x + 1\right) = 0$

$\left(9 {x}^{2}\right) \left(x + 1\right) \left(x + 1\right) = 0$

Since we know the product of the three terms on the left comes out to 0, we know that any one of those terms evaluating to 0 will be a solution of the equation. Thus, we solve each factor separately to determine all possible solutions for $x$:

$9 {x}^{2} = 0$

${x}^{2} = 0$

$x = \pm 0 \implies x = 0 \text{ (Multiplicity 2)}$

$x + 1 = 0 \implies x = - 1$

$x + 1 = 0 \implies x = - 1$

We get four roots: $x = - 1 , - 1 , 0 , 0$.