How to factorized this 4th degree polynomial?

Question

I need your help to this polynomial's factorization.

Factorize this polynomials which doesn't have roots in Q.

$ \ f(x) = x^4 +2x^3-8x^2-6x-1 $

P.S.) Are there any generalized method finding 4^th degree polynomials factor?

Answer 1

I am not sure what you mean about being irreducible over $\mathbb Q$ - it may have no roots over $\mathbb Q$, which is a different thing.

The general process for finding the roots of a quartic starts by trying to express it as a difference of two squares, and solving a cubic for the parameter involved (you can pair the four roots of the quartic in three different ways, which represent three different factorisations into two quadratics).

Often trying to spot a difference of two squares is a good method with quartics anyway, if you are stuck.

Here I spotted $x^4+2x^3+x^2-9x^2-6x-1$ as a possible way forward - and this leads to a factorisation into quadratics with integer coefficients.

Answer 2

You can always find the roots of a fourth degree polynomial. The formulas are not pretty, but they do exist.

Using the formulas, or, like me, using Wolfram Alpha, can show you that the roots of the polynomial are $1\pm\sqrt{2}$ and $-2\pm \sqrt{3}$ so the polynomial is

$$\begin{align}f(x)&=(x-1-\sqrt{2})(x-1+\sqrt{2})(x+2-\sqrt{3})(x+2+\sqrt{3})\\&=(x^2-2x-1)(x^2+4x+1)\end{align}$$

Answer 3

Yes, there is a general method for determining whether or not a quartic is irreducible in $\mathbb Q[x]$:

1. Does it have rational roots? In your case, the answer is negative, by the rational root theorem.
2. Consider the polynomial $q(x)=p\left(x-\frac12\right)=x^4-\frac{19 x^2}2+3x-\frac3{16}$.
3. Consider the resolvent cubic of $q(x)$:\begin{align}r(x)&=x^3+2\times\left(-\frac{19}2\right)x^2+\left(\left(-\frac{19}2\right)^2-4\times\left(-\frac3{16}\right)\right)x-3^2\\&=x^3-19x^2+91x-9.\end{align}
4. Does it have a rational root which is a perfect square in $\mathbb Q$? Yes: $9(=3^2)$ is such a root.
5. Then $q(x)=\left(x^2+3x+\beta\right)\left(x^2-3x+\gamma\right)$, with$$\beta=\frac12\left(-\frac{19}2+3^2-\frac33\right)=-\frac34\text{ and }\gamma=\frac12\left(-\frac{19}2+3^2+\frac33\right)=\frac14.$$
6. Since $p(x)=q\left(x+\frac12\right)$, you get a decomposition for $p(x)$:$$p(x)=(x^2+4x+1)(x^2-2x-1).$$

Answer 4

For all real $k$ we have:

$$x^2+2x^3-8x^2-6x-1=(x^2+x+k)^2-x^2-k^2-2kx^2-2kx-8x^2-6x-1=$$ $$=(x^2+x+k)^2-((2k+9)x^2+2(k+3)x+k^2+1).$$ Now, we'll choose a value of $k$ such that $$(2k+9)x^2+2(k+3)x+k^2+1=(ax+b)^2.$$ Easy to see that $k=0$ is valid.

In the general we need to solve the following equation. $$(k+3)^2-(2k+9)(k^2+1)=0.$$ Id est, $$x^2+2x^3-8x^2-6x-1=(x^2+x+k)^2-x^2-k^2-2kx^2-2kx-8x^2-6x-1=$$ $$=(x^2+x+k)^2-((2k+9)x^2+2(k+3)x+k^2+1)=$$ $$=(x^2+x)^2-(9x^2+6x+1)=(x^2+x)^2-(3x+1)^2=(x^2-2x-1)(x^2+4x+1).$$

Answer 5

Just working through the hint in the accepted answer, and verified with the Wolfram result in the second answer. Just for me, and maybe a check/guide for someone else:

!$x^4+2x^3+x^2−9x^2−6x−1 \\ (x^2+x)^2 - (3x+1)^2 \\ \\ ((x^2+x) + (3x+1)) ((x^2+x) - (3x+1)) \\ (x^2+4x+1) (x^2-2x-1) \\ \\ \text{quadratic formula for } (x^2+4x+1)\\ -4/2\pm\sqrt{16-4}/2 \\ -2 \pm\sqrt3 \\ \\ \text{quadratic formula for } (x^2-2x-1) \\ --2/2\pm\sqrt{4-4(-1)}/2 \\ 1\pm\sqrt{4+4}/2 \\ 1\pm\sqrt2 \\ \\ \text{factored form }\\ (x+2+\sqrt3)(x+2-\sqrt3)(x-1+\sqrt2)(x-1-\sqrt2)\\ $