# How do you factor the expression 4x^3 + 6x^2 + 6x + 9?

Dec 24, 2015

Factor by grouping to find:

$4 {x}^{3} + 6 {x}^{2} + 6 x + 9 = \left(2 {x}^{2} + 3\right) \left(2 x + 3\right)$

#### Explanation:

Factor by grouping:

$4 {x}^{3} + 6 {x}^{2} + 6 x + 9$

$= \left(4 {x}^{3} + 6 {x}^{2}\right) + \left(6 x + 9\right)$

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

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

The remaining quadratic factor $\left(2 {x}^{2} + 3\right)$ has no simpler linear factors with Real coefficients since it is non-zero for any Real number $x$:

$2 {x}^{2} + 3 \ge 3 > 0$

If we resort to Complex numbers then it factorises further:

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