factoring polynomials I need help with a challenge problem I'm attempting to solve in my math book.
The first one is:  Factor $4x^2(x-3)^3-6x(x-3)^2+4(x-3)$
I worked through the problem and got {$[x-3][4x^2(x-3)^2-6x(x-3)+4]$}.
Could someone please look at this and tell me if I'm correct, wrong, or even close to the right answer.
 A: You can pull out a $2$ and get
$$
2(x-3)(2x^2(x-3)^2 -3x(x-3)+2).
$$
If you then let $w=x(x-3)$, then your second factor is $2w^2-3w+2$.  The discriminant of that polynomial is $b^2-4ac=(-3)^2-4\cdot2\cdot2=-7$.  So if you allow complex numbers, you could factor this further.  Barring that, you can expand what you've got:
$$
\begin{align}
& \phantom{{}={}} 2x^2(x-3)^2 -3x(x-3)+2 \\[8pt]
& = 2x^2(x^2-6x+9) - (3x^2 - 9x) + 2 \\[8pt]
& = 2x^4 -12x^3+18x^2 - 3x^2 + 9x + 2 \\[8pt]
& = 2x^4 -12x^3 +15x^2 + 9x+2
\end{align}
$$
There's a criterion for whether this has rational roots.  If you need to factor using only integer coefficients, that finishes the problem off.  Otherwise, it might take a lot more work.
A: One can certainly go a bit farther in the direction that @MichaelHardy points us in, by setting $2w^2-3w+2=0$, whose roots are
$$
w=\frac{3\pm\sqrt{-7}}4\,.
$$
Then feed this into the equation $x(x-3)=w$, that is,
$$
x^2-3x-\frac{3\pm\sqrt{-7}}4=0\,,
$$
whose discdriminant now is $\Delta'=9-(3\pm\sqrt{-7}\,)=6\mp\sqrt{-7}$. The cognoscenti will notice that this number not only is an algebraic integer in the quadratic field $\mathbb Q(\sqrt{-7}\,)$, but since its norm is 43, it’s indecomposable there (because $h=1$). Thus there’s certainly no hope of getting a square root of $\Delta'$ in terms of $\sqrt{-7}$. Easy enough now to write down the roots of $2x^2(x-3)^3-3x(x-3)+2$, they’re
$$
\frac{3\pm\sqrt{6\pm\sqrt{-7}}}2\,.
$$
