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I have some quartic polynomial I wish to factor. Here is an example:

$x^4 + x^2 + 1 $

I know the answer to this question

$ (x^2 + x + 1)(x^2 -x +1) $

We get these 2 irreducible quadratics. I am trying to see if there is some systematic way, a process to arrive at this, without having to figure out a pattern for this.

Palu
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2 Answers2

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If you have this nice sort of symmetry there's a trick you can apply; divide through by $x^2, x\ne 0$ to get $$x^2 + 1 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 -1$$

Using the substitution $u=x + 1/x$ allows you to solve for the roots of the polynomial.

Something else you could do: take the polynomials $ax^2 + bx + c$ and $dx^2+ex + f$, multiply them together and solve for $a,b,c,d,e,f$. Note that you'll end up with 6 unknowns and 5 equations, so you'll have to choose one of the values (say $a=1$).

Quantaliinuxite
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  • Hi there, thanks for your reply. That is interesting, in terms of trying to go further with this to find the roots. But I was asking how to go thru a systematic process to get the quartic factored down to those 2 irreducible quadratics, if there is such a way. – Palu Aug 22 '18 at 23:48
  • OK Quanta, I think i briefly misunderstood you, i thought you were reducing the irreducible. I will try this method of yours, and see what i get. – Palu Aug 22 '18 at 23:52
  • Hi Quanta, I tried it out. That is quite an ingenious approach/trick. I will give you as the answer to my question, only because it is a clever trick. BUT i hope others will put another possible approach to this question as well. – Palu Aug 22 '18 at 23:56
  • I added some more, you might find it useful – Quantaliinuxite Aug 23 '18 at 00:17
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You can also use the differences of squares identity just as below: $$x^4+x^2+1=x^4+2x^2+1-x^2$$ The first three summands is conveniently $(x^2+1)^2$ and the equation can be rewritten into: $$(x^2+1)^2-x^2$$ and by differences of squares identity you would get your two quadratics.

MathEnthusiast
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  • That is really great MathEnthusiast!!! I wish I could vote for more than one solution as the solution. Both you and Quantaliinuxite, deserve to be the solutions to this problem i presented here, unfortunately, I can't assign 2 solutions. Thanks once again for this amazing solutions here. – Palu Aug 23 '18 at 05:27