How do you factor the expression $x^{4} + 6 x^{2} - 7$ $x^4 +6x ^2-7$ ?

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Answer 1 · 2016-11-29T00:32:43

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

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

$1 + 6 - 7 = 0$ $1+6-7 = 0$

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

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

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

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

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

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

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

How do you factor the expression x4+6x2−7x^4 +6x ^2-7?