How do i factor x^4+6x^2-7 completly?

The factoring can be done similarly to a quadratic equation thanks to x^4 being the square value of x^2.

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

Factored completely we get: (x^2 + 7)(x - 1)(x + 1)

Just make it into ax^2+bx+c form
remember that
(x^2)^2=x^4
(x^2)=x^2

rewriete in normal form (for people who are still understnain stuf)

(x^2)^2+6(x^2)-7
now find what 2 numbers multiply to get -7 and add to 6
numbers are -1 and 7
((x^2)-1)((x^2)+7)
(x^2-1)(x^2+7)
diffrence of 2 perfect squaer
a^2-b^2=(a-b)(a+b)
(x)^2-(1)^2=(x-1)(x+1)

totla factored form is
(x-1)(x+1)(x^2+7)