# I need help factoring quadratics.

I need help factoring $x^4-2x^2-3=0$. I do not know how to factor this disguised quadratic.

• @Chief123: What if you let $x^2 = w$?
– Moo
Feb 19 '16 at 4:12
• Yes we should substitute. @Moo Feb 19 '16 at 4:15
• Do you see how to solve using that hint? Try it.
– Moo
Feb 19 '16 at 4:16
• Yes I knew it was. I factored it to (u-3)(x+1)=0, u=x^2. I do not know what to do next, @SS_C4 Feb 19 '16 at 4:20
• Well, you now have $u = 3, -1$ and $x = \pm u^{1/2}$. So, $x = \pm \sqrt{3}, ~ \pm i$.
– Moo
Feb 19 '16 at 4:22

take $y=x^2$ hence you have $y^2-2y-3=0$ hance $(y-3)(y+1)=0$

• it's possible to simplify further to $(x^2-3)(x^2+1)=(x+\sqrt{3})(x-\sqrt{3})(x^2+1)$ Feb 19 '16 at 4:24
• i suppose if it's over reals Feb 19 '16 at 4:25

HINT

Rewrite $-2x^2$ as $-3x^2+x^2$ and group.

More specifically, $x^4-2x^2-3$ is equivalent to $x^4-3x^2+x^2-3$.

Can you proceed?

• it's $2x^2$ . Please edit. Feb 19 '16 at 4:17
• Done, thanks @WinVineeth Feb 19 '16 at 4:18