# How do you factor the expression 15x^3-23x^2+4x?

Jan 17, 2017

$15 {x}^{3} - 23 {x}^{2} + 4 x = x \left(5 x - 1\right) \left(3 x - 4\right)$

#### Explanation:

Separate out the common factor $x$, then use an AC method...

$15 {x}^{3} - 23 {x}^{2} + 4 x = x \left(15 {x}^{2} - 23 x + 4\right)$

To factor the quadratic, look for a pair of factors of $A C = 15 \cdot 4 = 60$ with sum $B = 23$.

The pair $20 , 3$ works.

Use this pair to split the middle term and factor by grouping:

$15 {x}^{2} - 23 x + 4 = 15 {x}^{2} - 20 x - 3 x + 4$

$\textcolor{w h i t e}{15 {x}^{2} - 23 x + 4} = \left(15 {x}^{2} - 20 x\right) - \left(3 x - 4\right)$

$\textcolor{w h i t e}{15 {x}^{2} - 23 x + 4} = 5 x \left(3 x - 4\right) - 1 \left(3 x - 4\right)$

$\textcolor{w h i t e}{15 {x}^{2} - 23 x + 4} = \left(5 x - 1\right) \left(3 x - 4\right)$

Putting it all together:

$15 {x}^{3} - 23 {x}^{2} + 4 x = x \left(5 x - 1\right) \left(3 x - 4\right)$