TITLE: Factorize $3m^4-6m^3+14m^2-6m+11$
QUESTION [6 upvotes]: I have this expression:
$3m^4-6m^3+14m^2-6m+11=0$ and I want to factorize it in $(m^2+1)(3m^2-6m+11)$.
How can I do it? Thanks for any help!

REPLY [9 votes]: Hint $\ $ Because the leading and constants coefficients are primes, the possible factors are highly constrainted, so we can quickly find quadratic factors by undetermined coefficients. First, notice $\rm\:mod\ 3\!:\ -f \equiv x^2+1,\:$ so we check for a factor of form $\rm\: x^2\! +\, 3a\, x + 1.\:$ Its cofactor must have leading coefficient $3$ and constant coefficient $11$, i.e.
$$\rm\begin{eqnarray} (x^2+3a\,x+1)(3\,x^2\!+b\,x+11) &\,=\:&\rm 3\,x^4 + (b\!+\!9a)\, x^3 + (14\!+\!3ab)\, x^2 + (b\!+\!33a)\,x + 11 \\
&=&\rm 3\, x^2 - 6\, x^3 + 14\, x^2 - 6\,x + 11\end{eqnarray}$$
Comparing $\rm\:x^2$ terms, $\rm\:14=14\!+\!3ab\Rightarrow ab=0\:$ so $\rm\,a=0\,$ or $\rm\,b=0.\:$ If $\rm\:b=0\:$ then comparing $\rm\,x^3$ terms, $\rm\:9a=-6,\,$ contra $\rm\:a\in \Bbb Z.\:$ Thus $\rm\:a=0.\:$ Comparing $\rm\,x^3$ terms, $\rm\:b = -6,\:$ which works.