# How do you factor completely x^4+3x^3+6x^2+12x+8=0?

Mar 30, 2018

Over the reals we have:

$\left(x + 1\right) \left(x + 2\right) \left({x}^{2} + 4\right) = 0$

Over the complex numbers we have:

$\left(x + 1\right) \left(x + 2\right) \left(x - 2 i\right) \left(x + 2 i\right) = 0$

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} + 3 {x}^{3} + 6 {x}^{2} + 12 x + 8$

Note that:

$f \left(- 1\right) = 1 - 3 + 6 - 12 + 8 = 0$

So $x = - 1$ is a zero and $x + 1$ a factor:

${x}^{4} + 3 {x}^{3} + 6 {x}^{2} + 12 x + 8 = \left(x + 1\right) \left({x}^{3} + 2 {x}^{2} + 4 x + 8\right)$

Note that in the remaining cubic, the ratio of the first to the second term is the same as that of the third to the fourth term. So this cubic will factor by grouping:

${x}^{3} + 2 {x}^{2} + 4 x + 8 = \left({x}^{3} + 2 {x}^{2}\right) + \left(4 x + 8\right)$

$\textcolor{w h i t e}{{x}^{3} + 2 {x}^{2} + 4 x + 8} = {x}^{2} \left(x + 2\right) + 4 \left(x + 2\right)$

$\textcolor{w h i t e}{{x}^{3} + 2 {x}^{2} + 4 x + 8} = \left({x}^{2} + 4\right) \left(x + 2\right)$

Note that the remaining quadratic is always positive for real values of $x$. So it has no real zeros and no linear factors with real coefficients.

So factored completely over the real numbers we have:

${x}^{4} + 3 {x}^{3} + 6 {x}^{2} + 12 x + 8 = \left(x + 1\right) \left(x + 2\right) \left({x}^{2} + 4\right)$

If we allow the use of complex numbers, then note that:

${\left(2 i\right)}^{2} = 4 {i}^{2} = - 4$

So:

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

So factored completely over the complex numbers we have:

${x}^{4} + 3 {x}^{3} + 6 {x}^{2} + 12 x + 8 = \left(x + 1\right) \left(x + 2\right) \left(x - 2 i\right) \left(x + 2 i\right)$