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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 Feb 18 '14 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$$=(a+b)(a^2-ab+b^2)$. $\endgroup$ – Lucian Feb 19 '14 at 2:18
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I would start by factoring $x^3$ out from the first two terms and noticing the pattern in the result. $$ \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 Feb 18 '14 at 17:28
  • $\begingroup$ Very nice! Without such tricks, kroneckers method is necessary to factor a polynomial. $\endgroup$ – Peter Feb 18 '14 at 17:29
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    $\begingroup$ @Surferonthefall you have to factor many pages of those and you develop intuition :-) $\endgroup$ – gt6989b Feb 18 '14 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 gregory Sep 28 '17 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)? $\endgroup$ – MCT Apr 15 '14 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 Apr 16 '14 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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