# How do you factor completely: 2x^3 + 14x^2 + 4x + 28?

Jul 24, 2015

Separate out the common scalar factor, then factor by grouping to find:

$2 {x}^{3} + 14 {x}^{2} + 4 x + 28 = 2 \left({x}^{2} + 2\right) \left(x + 7\right)$

#### Explanation:

$2 {x}^{3} + 14 {x}^{2} + 4 x + 28$

$= 2 \left({x}^{3} + 7 {x}^{2} + 2 x + 14\right)$

$= 2 \left(\left({x}^{3} + 7 {x}^{2}\right) + \left(2 x + 14\right)\right)$

$= 2 \left({x}^{2} \left(x + 7\right) + 2 \left(x + 7\right)\right)$

$= 2 \left({x}^{2} + 2\right) \left(x + 7\right)$

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

If you really want, you can factor ${x}^{2} + 2 = \left(x + i \sqrt{2}\right) \left(x - i \sqrt{2}\right)$

and hence:

$2 {x}^{3} + 14 {x}^{2} + 4 x + 28 = 2 \left(x + i \sqrt{2}\right) \left(x - i \sqrt{2}\right) \left(x + 7\right)$