Simplify $x^3 - 4x^2 + 10x - 125$ I've been trying to simplify $x^3 - 4x^2 + 10x - 125$ for a while now, and I don't seem to progress.
I know that the factors of $125$ are $1$, $5$, $25$ and $125$, but none of these seem to help here. So far I have managed to get $x(x^2 - 4x + 10) - 125$.
Can I go any further than this? Thank you! By the way... it's my first post here, so if I didn't provide something necessary - excuse me. 
 A: If you want to factorize it, it's irreducible over the rationals. However, you can get things like $x(x(x−4)+10)−125$, which technically aren't factorizations.
If you don't want to restrict yourself to rationals, you can write it as $(x-5.89793)(x^2+1.89793 x+21.1938)$. (The exact form is extremely complicated.)
A: If you have access to any computer algebra system, you can solve for the roots of a cubic equation. (If you don't have access to a computer algebra system then writing everything out this fully is a waste of time!)
$$x^3 - 4x^2 + 10x - 125=(x-x_0)(x-x_1)(x-x_2)=\left(x-\frac{1}{3} 7^{2/3} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{2}{449+15 \sqrt{897}}}+\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(449+15 \sqrt{897}\right)}-\frac{4}{3}\right) \left(x-\frac{1}{3} 7^{2/3} \left(1-i \sqrt{3}\right) \sqrt[3]{\frac{2}{449+15 \sqrt{897}}}+\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(449+15 \sqrt{897}\right)}-\frac{4}{3}\right) \left(x+\frac{1}{3} \left(2\ 7^{2/3} \sqrt[3]{\frac{2}{449+15 \sqrt{897}}}-\sqrt[3]{\frac{7}{2} \left(449+15 \sqrt{897}\right)}-4\right)\right)$$
