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I am trying to factor

$$x^5+4x^3+x^2+4=0$$

I've used Ruffini's rule to get

$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$

But I don't know what to do next.

The solution is $(x+1) (x^2+4) (x^2-x+1) = 0$. I've tried using the completing square method but with no result. Could you give me hints?

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    $\begingroup$ Since you know the solution just try to go backwards to the previous step $\endgroup$
    – Math137
    Commented Feb 18, 2014 at 17:23
  • $\begingroup$ Those two terms containing the number $4$ just beg to be grouped together and factored, wouldn't you agree ? Then, in the remaining two, $x^2$ is an obvious choice for a common factor. But now we have two groups of terms, each containing $x^3+1$ ! After factoring that, we notice that the remaining quadratic has no real roots, and $x^3+1$ can be factored by realizing that $1=1^3$, and using the fact that $a^3+b^3$$=(a+b)(a^2-ab+b^2)$. $\endgroup$
    – Lucian
    Commented Feb 19, 2014 at 2:18

4 Answers 4

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I would start by factoring $x^3$ out from the first two terms and noticing the pattern in the result. $$ \begin{split} x^5+4x^3 + x^2 + 4 &= x^3 \left(x^2+4\right) + x^2+4 \\ &= \left(x^2+4\right)\left(x^3+1\right) \\ &= \left(x^2+4\right)(x+1)\left(x^2-x+1\right), \\ \end{split} $$ where the last step is the standard factoring of the sum of two cubes.

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  • $\begingroup$ Great! How did you know it? :) $\endgroup$
    – Surfer on the fall
    Commented Feb 18, 2014 at 17:28
  • $\begingroup$ Very nice! Without such tricks, kroneckers method is necessary to factor a polynomial. $\endgroup$
    – Peter
    Commented Feb 18, 2014 at 17:29
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    $\begingroup$ @Surferonthefall you have to factor many pages of those and you develop intuition :-) $\endgroup$
    – gt6989b
    Commented Feb 18, 2014 at 17:29
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    $\begingroup$ Look at the coefficients! They are $1,4,1,4$. Sometimes it can get a bit more complicated. For example $2,-3,-6,9 = 2,-3,-3(2), -3(-3)$ $\endgroup$
    – Steven Alexis Gregory
    Commented Sep 28, 2017 at 15:26
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Or, alternatively, note that $$x^4-x^3+5x^2-4x+4=x^2(x^2-x+1)+4x^2-4x+4$$ and factor $4$ from the last three terms.

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  • $\begingroup$ Is there any motivation for that manipulation (that isn't reverse engineered)? $\endgroup$
    – MT_
    Commented Apr 15, 2014 at 21:48
  • $\begingroup$ @Michael T - Nothing other than learning at an early age (doing lots of integrals and other things) that it sometimes pays to split one term into the sum (or product sometimes) of two terms. Writing the coefficient on $x^2$ as $1+4$ creates a nice sequence of coefficients: 1, -1, 1; 4, -4, 4; and the rest just flows from there. As gt6989b said above, if you do enough of these things, you develop an intuition. Some of these things are as much art as science. $\endgroup$
    – Chris Leary
    Commented Apr 16, 2014 at 5:07
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Here I think this will work . Main idea being splitting up of $5x^2$ as $4x^2 + x^2$

$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$ $$(x+1) (x^4-x^3+4x^2+ x^2-4x+4)=0 $$ $$(x+1) (x^4+4x^2+ x^2-x^3-4x+4)=0 $$ $$(x+1)[ x^2(x^2+4)+ (1-x)( x^2 +4)]=0 $$ $$(x+1)[ (x^2+4)(x^2+1-x))=0 $$ $$(x+1)[ (x^2+4)(x^2-x+1))=0 $$

Hence proved

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The answers actually $$ (x^2 + 4)(x^3 + 1)\\ x^5 + 4x^3 + x^2 + 4\\ (x^5 + 4x^3)(x^2 + 4)\\ x^3(x^2 + 4) 1(x^2 + 4)\\ (x^2 + 4)(x^3 + 1) $$ (Sorry, too lazy to actually change them to exponents)

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