Trick for reducing a specific polynomial.

Question

Let $f(x) = x^4 − x^3 + 2x^2 − 3x − 3$

Show that f(x) is reducible over $\Bbb{Q}$

Rational zeros shows there is no roots i tried writing $(x^2+bx+c)(x^2+dx+e)$ and then tried to solve the equations for each of letters but it got crazy ugly which lead me to the conclusion there must be a better way...

Answer 1

i think that is a good idea, it is $$(x^2+3)(x^2-x-1)=f(x)$$

Answer 2

Also, we can use the following way.

For all real $k$ we obtain: $$x^4-x^3+2x^2-3x-3=\left(x^2-\frac{1}{2}x+k\right)^2-\frac{1}{4}x^2-k^2-2kx^2+kx+2x^2-3x-3=$$ $$=\left(x^2-\frac{1}{2}x+k\right)^2-\left(\left(2k-\frac{7}{4}\right)x^2-(k-3)x+k^2+3\right).$$ Now we'll choose a value of $k$ such that $2k-\frac{7}{4}>0$ and $$(k-3)^2-4\left(2k-\frac{7}{4}\right)(k^2+3)=0.$$ Easy to see that $k=1$ is valid.

Id est, $$x^4-x^3+2x^2-3x-3=\left(x^2-\frac{1}{2}x+1\right)^2-\left(\frac{1}{4}x^2+2x+4\right)=$$ $$=\left(x^2-\frac{1}{2}x+1\right)^2-\left(\frac{1}{2}x+2\right)^2=(x^2-x-1)(x^2+3).$$