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

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  • $\begingroup$ @Chief123: What if you let $x^2 = w$? $\endgroup$
    – Moo
    Feb 19 '16 at 4:12
  • $\begingroup$ Yes we should substitute. @Moo $\endgroup$
    – Chief123
    Feb 19 '16 at 4:15
  • $\begingroup$ Do you see how to solve using that hint? Try it. $\endgroup$
    – Moo
    Feb 19 '16 at 4:16
  • $\begingroup$ 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 $\endgroup$
    – Chief123
    Feb 19 '16 at 4:20
  • $\begingroup$ Well, you now have $u = 3, -1$ and $x = \pm u^{1/2}$. So, $x = \pm \sqrt{3}, ~ \pm i$. $\endgroup$
    – Moo
    Feb 19 '16 at 4:22
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take $y=x^2$ hence you have $y^2-2y-3=0$ hance $(y-3)(y+1)=0$

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    $\begingroup$ it's possible to simplify further to $(x^2-3)(x^2+1)=(x+\sqrt{3})(x-\sqrt{3})(x^2+1)$ $\endgroup$
    – Itakura
    Feb 19 '16 at 4:24
  • $\begingroup$ i suppose if it's over reals $\endgroup$
    – user300
    Feb 19 '16 at 4:25
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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?

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

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