NO, you have not factored anything at all
x^4 - 6x^2 + 5 = 0
(x^2 - 1)(x^2 - 5) = 0
(x-1)(x+1)x^2 - 5) = 0 if you factor over the rationals
(x-1)(x+1)x-√5)(x+√5) = 0 if you factor over the reals
start by treating the problem like it is a quadratic. Let u = x^2
u^2 - 6u + 5
Factors as: (u-5)(u-1)
(x^2 -5)(x^2 -1) =0
Whenever there is = 0, Normally, you have to solve for x and not stop at the factoring step.
Set each factor = to zero.
(x^2-5)=0 or (x^2 - 1) = 0
x^2 = 5 or x^2 = 1
x = square root of 5 divided by 2
x = minus the square root of 5 divided by 2
x = 1
x = -1
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