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So I was thinking that how can I factorize $x^4-2x^2-8$, or any $4$ degree polynomial, without going for the first factor by trial and error method?

José Carlos Santos
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Asad
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  • There are formulas for quartics but they're too unwieldy and unstable. – Parcly Taxel Jul 19 '17 at 05:27
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    There is in fact a (very lengthy and tedious) general form you can use for solving for the roots of fourth-degree polynomials but noone learns or memorizes it due to its incredibly unwieldy nature. In your very specific case however, there is a nice way to simplify the problem first and use what you know about quadratics by replacing $x^2$ by something else, lets call it $u$. So, $x^4-2x^2-8=(x^2)^2-2(x^2)-8=u^2-2u-8$. Now, you can solve for $u$. Given the solutions for $u$, you can then go back and replace with $u=x^2$ again and solve for $x$. – JMoravitz Jul 19 '17 at 05:28
  • As an aside, it has been proven thanks to Galois Theory that there does not and can not exist a general solution for the roots of arbitrary degree five or higher polynomials (though as usual, special cases occur and can have very clean answers). Link for arbitrary degree 4 solution. – JMoravitz Jul 19 '17 at 05:30
  • If you need an explicit formula, check out the third answer on this post https://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations-quartic Hope it helps :) – Jonathan Davidson Jul 19 '17 at 05:32

3 Answers3

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$$x^4-2x^2-8=x^4-2x^2+1-3^2=(x^2-4)(x^2+2)=(x-2)(x+2)(x^2+2).$$

We can make factorization for all $4$ degree polynomial over $\mathbb R$ by the Ferrari's method.

Michael Rozenberg
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Express the function as a quadratic in $x^2$

$(x^2)^2 - 2x^2 -8$

Factor it for $x^2$

$(x^2+2)(x^2-4)$

Write 4 as $2^2$ and expand it,

$(x^2+2)(x+2)(x-2)$

Coderzz
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Let $x^2=y$.

Then you have to factorize $y^2-4y-8=P(y)$

$P(y)=(y-2- \sqrt{12})(y-2+ \sqrt{12})$

Thus $$x^4-4x^2-8=(x^2-2- \sqrt{12})(x^2-2+ \sqrt{12})=$$ $$(x+ \sqrt{2+ \sqrt{12}})(x-\sqrt{2+ \sqrt{12}})((x^2-2+ \sqrt{12})$$

Marios Gretsas
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