# How do you factor completely  2x^3 + 4x^2 + 6x + 12?

Dec 31, 2017

$\left({x}^{2} + 3\right) \left(2 x + 4\right)$ or $\left(x + i \sqrt{3}\right) \left(x - i \sqrt{3}\right) \left(2 x + 4\right)$

#### Explanation:

$y = 2 {x}^{3} + 4 {x}^{2} + 6 x + 12$

Moving $6 x$ closer to $2 {x}^{3}$

$y = 2 {x}^{3} + 6 x + 4 {x}^{2} + 12$

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

Take out ${x}^{2} + 3$

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

we can continue factoring ${x}^{2} + 3$ using complex numbers: ${i}^{2} = - 1$

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

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

$y = \left(x + i \sqrt{3}\right) \left(x - i \sqrt{3}\right) \left(2 x + 4\right)$

Dec 31, 2017

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

#### Explanation:

$\text{take out "color(blue)"common factor of 2}$

$\Rightarrow 2 \left({x}^{3} + 2 {x}^{2} + 3 x + 6\right)$

$\text{when } x = - 2 \to {x}^{3} + 2 {x}^{2} + 3 x + 6 = 0$

$\Rightarrow \left(x + 2\right) \text{ is a factor}$

$\textcolor{red}{{x}^{2}} \left(x + 2\right) \cancel{\textcolor{m a \ge n t a}{- 2 {x}^{2}}} \cancel{+ 2 {x}^{2}} + 3 x + 6$

$= \textcolor{red}{{x}^{2}} \left(x + 2\right) \textcolor{red}{+ 3} \left(x + 2\right) \cancel{\textcolor{m a \ge n t a}{- 6}} \cancel{+ 6}$

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

$\text{factor } {x}^{2} + 3$

${x}^{2} + 3 = 0 \Rightarrow {x}^{2} = - 3 \Rightarrow x = \pm \sqrt{3} i$

$\Rightarrow 2 {x}^{3} + 4 {x}^{2} + 6 x + 12$

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