# How do you factor completely x^3-5x^2+6x?

##### 1 Answer
Jun 5, 2016

To factor ${x}^{3} - 5 {x}^{2} + 6 x$ completely
$x \left(x - 2\right) \left(x - 3\right)$

#### Explanation:

To factor ${x}^{3} - 5 {x}^{2} + 6 x$ completely

Begin by factoring out the $x$ common to each term

$x \left({x}^{2} - 5 x + 6\right)$

Next find the factors of the third term $6$

1x6 and 2x3

Since the second sign is $+$ the factors must add up to the middle term $- 5$. We will use $- 2 \mathmr{and} - 3$

Now factor the trinomial

$x \left(x - 2\right) \left(x - 3\right)$