How do you factor completely: 4x^3 + 12x^2 + 3x + 9?

Jul 11, 2015

$4 {x}^{3} + 12 {x}^{2} + 3 x + 9 = \left(4 {x}^{2} + 3\right) \left(x + 3\right)$

This has no simpler linear factors with real coefficients, since $4 {x}^{2} + 3 \ge 3 > 0$ for all $x \in \mathbb{R}$.

Explanation:

$4 {x}^{3} + 12 {x}^{2} + 3 x + 9$

$= \left(4 {x}^{3} + 12 {x}^{2}\right) + \left(3 x + 9\right)$

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

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

The remaining quadratic factor $\left(4 {x}^{2} + 3\right)$ cannot be factored into linear factors with real coefficients, since $4 {x}^{2} + 3 \ge 3 > 0$ for all $x \in \mathbb{R}$.