# How do you factor y= 3x^7 - 9x ^6 + 18x ^5?

Dec 16, 2015

$y = 3 {x}^{7} - 9 {x}^{6} + 18 {x}^{5} = 3 {x}^{5} \left({x}^{2} - 3 x + 6\right)$

#### Explanation:

First note that all the terms are divisible by $3 {x}^{5}$, so separate that out as a factor:

$y = 3 {x}^{7} - 9 {x}^{6} + 18 {x}^{5} = 3 {x}^{5} \left({x}^{2} - 3 x + 6\right)$

The remaining quadratic factor is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 3$ and $c = 6$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - \left(4 \times 1 \times 6\right) = 9 - 24 = - 15$

Since this is negative, the quadratic has only Complex zeros and cannot be factored any further with Real coefficients.