# How do you factor h(x)=x^3-3x^2-x+3?

Dec 19, 2015

Factor by grouping then using the difference of squares identity to find:

${x}^{3} - 3 {x}^{2} - x + 3 = 0 = \left(x - 1\right) \left(x + 1\right) \left(x - 3\right)$

#### Explanation:

The difference of squares identity may be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

After factoring by grouping use this identity with $a = x$ and $b = 1$...

${x}^{3} - 3 {x}^{2} - x + 3$

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

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

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

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

$= \left(x - 1\right) \left(x + 1\right) \left(x - 3\right)$