# How do you factor 6x^3-6x^2+18x?

Oct 1, 2015

$6 x \left({x}^{2} - x + 3\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 6 {x}^{3} - 6 {x}^{2} + 18 x$

Extract the obvious common factor $6 x$ from each term:
$\textcolor{w h i t e}{\text{XXX}} 6 x \left({x}^{2} - x + 3\right)$

We could try to factor $\left({x}^{2} - x + 3\right)$
but an examination of the discriminant (see below) gives us
$\textcolor{w h i t e}{\text{XXX}}$Delta = (-1)^2-4(1)(3) < 0
which tells us there are no Real roots;
so factors are not available.

Discriminants and their significance
$\textcolor{w h i t e}{\text{XXX}}$for a quadratic in the form
$\textcolor{w h i t e}{\text{XXXXX}} a {x}^{2} + b x + c$
$\textcolor{w h i t e}{\text{XXX}}$the discriminant is
$\textcolor{w h i t e}{\text{XXXXX}} \Delta = {b}^{2} - 4 a c$

color(white)("XXX")Delta { (< 0 rarr "no Real roots"), (= 0 rarr "exactly one Real root"), (> 0 rarr "two Real roots"):}#