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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.

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You can still factor a constant out of $2$. – Amzoti Aug 14 '13 at 2:47
I would expand what you have between the second pair of brackets. – Lubin Aug 14 '13 at 2:47
up vote 3 down vote accepted

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.

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No. If you allow complex number, you could factor this further. – Michael Hardy Aug 14 '13 at 3:02
Typo. There was a missing part of a word. – Michael Hardy Aug 14 '13 at 3:07
You have an unbalanced parenthesis on the first line. – chubakueno Aug 14 '13 at 3:08
Also, wolfram spits a "nope" for reals. (…) – chubakueno Aug 14 '13 at 3:20

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 $$ \frac{3\pm\sqrt{6\pm\sqrt{-7}}}2\,. $$

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