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

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Answer 1 · 2016-11-29T00:32:43

#x^4+6x^2-7 = (x-1)(x+1)(x^2+7)#

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 #(x-1)#, #(x+1)# and their product #(x^2-1)# are all factors:

#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)#

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