# Finding the roots $x^4-4x^3-x^2-8x+4=0$ (contest math)

So the problem is :

$x^4-4x^3-x^2-8x+4=0$, find all solutions

A tip that I have gotten, is to divide both sides by $x^2$. I've tried so, but I do not manage to see any further. Do anyone know how this tip could help me?

(Yes, I'm aware that the polynomial above can be factorized into two degree 2 polynomials, which promptly gives me the answer. But that factorization would be extremely hard to spot, which is why I'm asking about the dividing)

Edit: meant to write $x^2$, not $2$

• If you want to learn how to solve any general quartic polynomial, check out this page on solving cubics and quartics. Jul 9 '16 at 12:49
• Dividing by $x^2$ is useless here Jul 9 '16 at 12:51
• @AmerYR As one of the answers shows, it's not useless. It gives you a quadratic easily factorable polynomial in terms of $x+\frac{2}{x}$. Jul 10 '16 at 7:44
• I saw that but I did not delete my comment. . I posted the comment before he answered it Jul 10 '16 at 10:39
• @NobleMushtak That just doesn't seem like the best idea when coming into a contest-math factorization problem, the roots are probably findable another way (probably easier) Jul 10 '16 at 11:45

Divide $x^4-4x^3-x^2-8x+4=0$ with $x^2$ in order to get the following $x^2-4x-1-\frac{8}{x}+\frac{4}{x^2}=(x+\frac{2}{x})^2-4(x+\frac{2}{x})-5=(x+\frac{2}{x}-5)(x+\frac{2}{x}+1)$. Finally, result is $(x^2-5x+2)(x^2+x+2)$.

• Perfect. @OP, it is made more clear by using $u = x+ \frac{2}{x}$ so that you get $u^2 - 4u - 5$, which is an easily factorisable quadratic. Jul 9 '16 at 12:59
• Ahh I see now, thank you Jul 9 '16 at 13:00

$x^4 - 4x^3 - x^2 -8 x + 4$ to try to factor it as a product of two polynomials with degree two I will try this

$x^4 -4x^3 -x^2-8x+4=(x^2+ax+c)(x^2+dx+e)$ but the constant term is 4 so we have two choices $c=1, e=4$ or $c=2, e=2$ if you choose the second you get the equations $a+d=-4, 4+ad=-1, 2a+2d=-8$ if you solve them you come up with a solution or maybe there is not a solution.

• You're assuming the constant terms are naturals, though. Jul 9 '16 at 13:13
• @ZainPatel: this is what they call an educated guess. If it works, it works.
– user65203
Jul 9 '16 at 13:14

Factor as $$(x^2-5x+2)(x^2+x+2)=0$$

• Well, the original poster had already said that he could factor it but was not asking about that! Jul 9 '16 at 12:51
• Have you carefully looked at the question? Jul 9 '16 at 12:52
• I think OP is more interested in how to derive/spot such factorizations rather than the actual factorization itself. Jul 9 '16 at 12:52
• The problem with your answer is, that it is "impossible" to see this factorisation for someone who do not know, how to solve these equations in general. Jul 9 '16 at 12:55
• All right guys, hypothesizing that the factorization is with integer coefficients, you know that the leading coefficients are $1$ and the constant terms multiplies to $4$. One of $(x^2+px+1)(x^2+qx+4)$ or $(x^2+px+2)(x^2+qx+2)$ or $(x^2+px+4)(x^2+qx+1)$, with $p+q=-4$.
– user65203
Jul 9 '16 at 13:09