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

Oct 12, 2015

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

#### Explanation:

Your starting expression looks like this

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

Notice that you can group the term according to their coefficients to get

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

Use $3$ and $2 x$ as common factors for their respective terms to get

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

Notice that you're dealing with the difference of two cubes, for which you know that

$\textcolor{b l u e}{{a}^{3} - {b}^{3} = \left(a - b\right) \cdot \left({a}^{2} + a b + {b}^{2}\right)}$

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

The expression can thus be written as

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

Use $\left(x - 1\right)$ as a common factor

$\left(x - 1\right) \cdot \left[3 \cdot \left({x}^{2} + x + 1\right) + 2 x\right]$

Finally, expand the paranthesis and group like terms to get

$\left(x - 1\right) \cdot \left(3 {x}^{2} + 3 x + 3 + 2 x\right) = \textcolor{g r e e n}{\left(x - 1\right) \left(3 {x}^{2} + 5 x + 3\right)}$