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How would you go about finding the roots of the polynomial: $$x^4 +5x^3+4x^2+6x-4=0.$$

I attempt to form two quadratics e.g. $$(x^2+ax+b)(x^2+cx+d)$$ and then tried to expand, collect like terms, and solve for the coefficients simultaneously, but this results in a hard system of equations.

I was wondering if there was a better approach to this question.

Any help would be greatly appreciated.

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    $\begingroup$ Rational Root Test? See: example: math.stackexchange.com/questions/12787/… $\endgroup$ – Moo Jan 31 at 22:20
  • $\begingroup$ Not sure whether Kronecker's method or factoring modulo some primes gives a faster solution. $\endgroup$ – Peter Jan 31 at 22:20
  • $\begingroup$ Maybe you try all the possible pairs of $b$ and $d$ multiplying to $-4$ ($b$ and $d$ must be integers). For every such pair, try to find $a$ and $c$. $\endgroup$ – Peter Jan 31 at 22:22
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Note that $x^4-4=(x^2+2)(x^2-2)$. This suggests that you might try to express your polynomial as$$(x^2+ax+2)(x^2+bx-2)=x^4+(a+b)x^3+abx^2+2(-a+b)x-4.$$It is now easy to see that all it takes is to choose $a=1$ and $b=4$.

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    $\begingroup$ I don't understand the initial remark. If you have $x^4+2x^2-5x-4$ then your suggestion is not very useful because the factorization is $(x^2+x+4)(x^2-x-1)$. $\endgroup$ – Robert Z Jan 31 at 22:36
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    $\begingroup$ Indeed. But what I wrote here corresponds to the way I approached the problem. In this case, it worked. If it had not worked, then I might have tried the general method. $\endgroup$ – José Carlos Santos Jan 31 at 22:43
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    $\begingroup$ Really? I think you were very lucky with your first approach. $\endgroup$ – Robert Z Jan 31 at 22:46
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    $\begingroup$ Does this answer boil down to: "Assume there are nice integer coefficients somewhere: that make things easier to brute force."? In general, this is not so simple a task. $\endgroup$ – drjpizzle Feb 1 at 1:32
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    $\begingroup$ I tried it this way (after trying to find a rational root) and in this specific case it worked. It's not deeper than that. $\endgroup$ – José Carlos Santos Feb 1 at 7:16
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Since there are no rational roots (a rational root should be a divisor of $4$), your attempt of factorization as $$x^4 +5x^3+4x^2+6x-4=(x^2+ax+b)(x^2+cx+d)$$ is a good idea.

Hint. By expanding the right-hand side and by comparing it with the the left-hand side, we have that $bd=-4$. Assuming that $b$ and $d$ are integers, by symmetry, you can try the following couples: $$(b,d)\in\{(4,-1), (2,-2), (-4,1)\}$$ and then solve the corresponding systems with respect to the remaining coefficients $c$ and $d$.

P.S. If you are lucky you will start with the couple $(2,-2)$. Be careful though, there are polynomials, which are "similar" to the given one, such that the other couples work:

i) $(4,-1)$ for $x^4+2x^2-5x-4=(x^2+x+4)(x^2-x-1)$.

ii) $(-4,1)$ for $x^4-4x^2+5x-4=(x^2+x-4)(x^2-x+1)$.

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I like the following way.

For all real value of $k$ we have:

$$x^4+5x^3+4x^2+6x-4=\left(x^2+\frac{5}{2}x+k\right)^2-\left(\left(2k+\frac{9}{4}\right)x^2+(10k-6)x+k^2+4\right).$$ Now, we'll choose a value of $k$, for which $2k+\frac{9}{4}>0$ and $$(5k-3)^2-\left(2k+\frac{9}{4}\right)(k^2+4)=0.$$ Easy to see that $k=0$ is valid.

Thus,$$x^4+5x^3+4x^2+6x-4=\left(x^2+\frac{5}{2}x\right)^2-\left(\frac{9}{4}x^2-6x+4\right)=$$ $$=\left(x^2+\frac{5}{2}x\right)^2-\left(\frac{3}{2}x-2\right)^2=(x^2+x+2)(x^2+4x-2).$$ Can you end it now?

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