Question

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

Answer 1

`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:

`9x^4 + 18x^3 + 9x^2 = 0`

`(9x^2)(x^2 + 2x + 1) = 0`

Now factor the quadratic:

`(9x^2)(x+1)(x+1) = 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`:

`9x^2 = 0`

`x^2 = 0`

`x = +- 0 => x = 0 " (Multiplicity 2)"`

`x+1 = 0 => x = -1`

`x + 1 = 0 => x = -1`

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