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

Dec 22, 2015

Find: $18 {x}^{3} + 9 {x}^{5} - 27 {x}^{2} = 9 {x}^{2} \left(x - 1\right) \left({x}^{2} + x + 3\right)$

as shown below...

#### Explanation:

Rearrange in standard order (descending powers of $x$) and separate out the common factor $9 {x}^{2}$ which all the terms are divisible by:

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

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

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

Next note that the sum of the coefficients of ${x}^{3} + 2 x - 3$ is $0$, so $x = 1$ is a zero of this cubic and $\left(x - 1\right)$ is a factor:

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

The discriminant of the remaining quadratic factor is ${1}^{2} - \left(4 \times 1 \times 3\right) = - 11$ which is negative, so there are no simpler factors with Real coefficients.

If you still want to factor it further you can use Complex coefficients:

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

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