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I have used many mathematical trick to factorise this polynomial : $4x^4-4x^3+4x^2+2x+1$ with real coeffecients but i didn't succeed because as I see all it's root are complex , and I want if there is any suitable method to factorise it ? and thanks in advanced .

Note: The motivation of this question is to compute some integral

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  • $\begingroup$ If all the roots are complex, then for sure I wouldn't look into factorization of polynomials that have odd degree. But even degrees might be worth looking into. So if you have $(ax^2+bx+c)(dx^2+ex+f)$, the question then becomes: Are there suitable coefficients $a,b,c,d,e,f$ that work? $\endgroup$ – imranfat May 27 '18 at 23:51
  • $\begingroup$ yes and that is what i want $\endgroup$ – zeraoulia rafik May 27 '18 at 23:55
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Let $$ x = \frac{i t }{ \sqrt 2} $$

Your polynomial is now $$ t^4 + i \sqrt 2 t^3 - 2 t^2 + i \sqrt 2 t + 1 $$ Divide by $t^2$ and then introduce $w = t + \frac{1}{t}$

dividing by $t^2$ $$ t^2 + i \sqrt 2 t - 2 + \frac{i \sqrt 2}{t} + \frac{1}{t^2} $$ which is $$ w^2 + i \sqrt 2 w - 4 $$

so the roots are $$ w = \frac{- i \sqrt 2 \pm \sqrt{14}}{2} $$ We get back to four roots in $t$ with two quadratics, $$ t^2 + \frac{ i \sqrt 2 \pm \sqrt{14}}{2} t + 1 = 0 $$

Finally we get four $x$ roots in conjugate pairs giving real quadratics as $(x - r)(x - \bar{r})$

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  • $\begingroup$ Beautiful! But are there any methods of guessing how to obtain such symmetry? $\endgroup$ – Przemysław Scherwentke May 28 '18 at 0:22
  • $\begingroup$ @PrzemysławScherwentke thank you. No method, really. First I tried $x=t/2$ I think. Then $x = i t/2$ to get the $\pm$ signs in a better pattern. Finally it seemed I needed $x = i t / \sqrt 2.$ You could say it was trial and error. $\endgroup$ – Will Jagy May 28 '18 at 0:30

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