# How do you factor y=18x^3 + 9x^5 - 27x^2 ?

Jan 2, 2016

$y = 18 {x}^{3} + 9 {x}^{5} - 27 {x}^{2}$

$= 9 {x}^{2} \left(x - 1\right) \left({x}^{2} + x + 3\right)$

$= 9 {x}^{2} \left(x - 1\right) \left(x + \frac{1}{2} - \frac{\sqrt{11}}{2} i\right) \left(x + \frac{1}{2} + \frac{\sqrt{11}}{2} i\right)$

#### Explanation:

First note that all of the terms are divisible by $9 {x}^{2}$, so separate out that common factor first.

$y = 18 {x}^{3} + 9 {x}^{5} - 27 {x}^{2}$

$= 9 {x}^{2} \left({x}^{3} + 2 x - 3\right)$

Then notice that the coefficients of the remaining cubic factor add up to $0$ (that is $1 + 2 - 3 = 0$), so $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

$= 9 {x}^{2} \left(x - 1\right) \left({x}^{2} + x + 3\right)$

The remaining quadratic factor has negative discriminant:

$\Delta = {b}^{2} - 4 a c = {1}^{2} - \left(4 \times 1 \times 3\right) = 1 - 12 = - 11$

So it has no factors with Real coefficients, but it can be factored using Complex coefficients that we can find using the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 1 \pm i \sqrt{11}}{2}$

So:

$\left({x}^{2} + x + 3\right) = \left(x + \frac{1}{2} - \frac{\sqrt{11}}{2} i\right) \left(x + \frac{1}{2} + \frac{\sqrt{11}}{2} i\right)$