$a^2 - 11ab + 3ab^2$
I tried to follow the usual method of factorization and got stuck. I rewrote the terms to get this:
$a^2- 11ba + 3b^2 \cdot a$
$11b$ and $3b^2$ are the ones that I need to split, right?
$a^2 - 11ab + 3ab^2$
I tried to follow the usual method of factorization and got stuck. I rewrote the terms to get this:
$a^2- 11ba + 3b^2 \cdot a$
$11b$ and $3b^2$ are the ones that I need to split, right?
$$a^2 - 11ab + 3ab^2 \\ = 3a(b^2-\frac {11}3b+\frac 13a)$$
The roots of $b^2-\frac {11}3b+\frac 13a$ can be obtained by the quadratic formula: $b_{1,2} = \frac {\frac {11}3\pm \sqrt{\frac {121}9 -\frac 43a}}{2} = \frac {11\pm \sqrt{121 -12a}}{6}$.
So you factor $a^2 - 11ab + 3ab^2$ as $\require{enclose}\enclose{box}{3a\left(b-\frac {11+ \sqrt{121 -12a}}{6}\right)\left(b-\frac {11- \sqrt{121 -12a}}{6}\right)}$
If you're not so much interested in factoring as getting it into an easier (and not potentially complex-valued) form, you could use completing the square to simplify $b^2-\frac {11}3b+\frac 13a$:
$$b^2-\frac {11}3b+\frac 13a \\ = \left(b-\frac{11}6\right)^2-\frac{121}{36}+\frac 13a \\ = \left(b-\frac{11}6\right)^2+\frac {12a-121}{36} \\ = \frac 1{36}\left[(6b-11)^2+12a-121\right]$$
Therefore $$a^2 - 11ab + 3ab^2 = \require{enclose}\enclose{box}{\frac 1{12}a\left[(6b-11)^2+12a-121\right]}$$