# How do you factor the expression x^4 +6x ^2-7?

Nov 29, 2016

${x}^{4} + 6 {x}^{2} - 7 = \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 7\right)$

#### Explanation:

Notice that the sum of the coefficients is $0$. That is:

$1 + 6 - 7 = 0$

Hence $x = 1$ is a zero. Also since all of the terms are of even degree, $x = - 1$ is also a zero.

So $\left(x - 1\right)$, $\left(x + 1\right)$ and their product $\left({x}^{2} - 1\right)$ are all factors:

${x}^{4} + 6 {x}^{2} - 7 = \left({x}^{2} - 1\right) \left({x}^{2} + 7\right) = \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 7\right)$

The remaining quadratic factor has no linear factors with Real coefficients.

If you use Complex numbers then it can be factored as:

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

but I would guess that you do not want to do that at Algebra 1 level.