Factorize $3m^4-6m^3+14m^2-6m+11$ 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!

You already factorized the  $3m^4 −6m^3 +14m^2 −6m+11$ with $(m^2 +1)(3m^2 −6m+11)$.
First factor equals to $(m^2 +1)$ and second $(3m^2 −6m+11)$.
$(m^2 +1)(3m^2 −6m+11) = 3m^4-6m^3+11m^2+3m^2-6m+11 = 3m^4-6m^3+14m^2+11$

Note:
But if you need solve $3m^4 −6m^3 +14m^2 −6m+11 = 0$ using factorization $(m^2 +1)(3m^2 −6m+11)$ it's a little different problem. $(m^2 +1)(3m^2 −6m+11) = 0$ is true if 1.:$m^2 +1 = 0$ or 2.:$3m^2 −6m+11=0$.
1. $m^2 +1 = 0$ => $m_{1,2} = \pm\sqrt{-1} = \pm{i}$
2. $3m^2 −6m+11=0$ => $m_{3,4} = 1\pm{2}i\sqrt{\frac{2}{3}}$
Here are four solutions:
1. $m = i$ : $3i^4-6i^3+14i^2-6i+11 = 3+6i+14-6i+11 = 0$
2. $m = -i$ : $3(-i)^4-6(-i)^3+14(-i)^2-6(-i)+11$$ = 3(-1)^4i^4-6(-1)^3i^3+14(-1)^2i^2-6(-i)+13 = 3 -6i-14+6i+13 = 0$
3. $m = 1+{2}i\sqrt{\frac{2}{3}}$: substitute to equation as in 1. or 2.
4. $m = 1-{2}i\sqrt{\frac{2}{3}}$: substitute to equation as in 1. or 2.
