<s>[INST] Factoring $(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$ and use the result to solve an equation I managed to prove that $(x+a+b+c)$ is a factor of
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$$
Then I was asked to use the result to solve
$$(x+2)(x-3)(x-1)+4=0$$
I know by comparison, $a=2, b=-3, c=-1$, and thus $(x-2)$ is a factor, but I can't really figure out how to solve the equation without expanding the brackets.
 [/INST] Probably not more efficient than straightforwardly factoring out, but it works as well:
$$\begin{eqnarray}\frac{(x+2)(x−3)(x−1)+4}{x-2} & = & \frac{(x-2+4)(x−3)(x−1)}{x-2}+\frac{4}{x-2} \\
& = & (1+\frac{4}{x-2})(x−3)(x−1)+\frac{4}{x-2} \\
& = & (x−3)(x−1) + \frac{4(x−3)(x−2+1)}{x-2}+\frac{4}{x-2}\\
& = & (x−3)(x−1) + 4(x−3)(1+\frac{1}{x-2})+\frac{4}{x-2}\\
& = & (x−3)(x−1) + 4(x−3) + \frac{4(x−2-1)}{x-2}+\frac{4}{x-2}\\
& = & (x−3)(x−1) + 4(x−3) + 4\\
& = & (x−3)(x+3) + 4\\
& = & x^2 - 5 \; .
\end{eqnarray}$$
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