# 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. Commented Jul 9, 2016 at 12:49
• Dividing by $x^2$ is useless here Commented Jul 9, 2016 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 $x+\frac{2}{x}$. Commented Jul 10, 2016 at 7:44
• I saw that but I did not delete my comment. . I posted the comment before he answered it Commented Jul 10, 2016 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) Commented Jul 10, 2016 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. Commented Jul 9, 2016 at 12:59
• Ahh I see now, thank you Commented Jul 9, 2016 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. Commented Jul 9, 2016 at 13:13
• @ZainPatel: this is what they call an educated guess. If it works, it works.
– user65203
Commented Jul 9, 2016 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! Commented Jul 9, 2016 at 12:51
• Have you carefully looked at the question? Commented Jul 9, 2016 at 12:52
• I think OP is more interested in how to derive/spot such factorizations rather than the actual factorization itself. Commented Jul 9, 2016 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. Commented Jul 9, 2016 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
Commented Jul 9, 2016 at 13:09