# How do you find all the real and complex roots of x^4 + 3x^2 + 2 = 0?

Feb 6, 2016

$x = \pm i , \pm i \sqrt{2}$

#### Explanation:

Note that ${x}^{2} + 3 x + 2$ can be factored into $\left(x + 1\right) \left(x + 2\right)$, so

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

So, we have the equation

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

Set both of these equal to $0$:

${x}^{2} + 1 = 0$

${x}^{2} = - 1$

$x = \pm i$

and

${x}^{2} + 2 = 0$

${x}^{2} = - 2 i$

$x = \pm i \sqrt{2}$

The equation has $4$ complex roots and no real roots (meaning it never crosses the $x$-axis:

graph{x^4+3x^2+2 [-13.83, 14.65, -2.31, 11.93]}