Resolve each of the following quadratic trinomial into factor:

The given expression is $6 \mathrm{a}^{2}+17 \mathrm{ab}-3 \mathrm{~b}^{2} .          \quad$ (Coefficient of $\mathrm{a}^{2}=6$, coefficient of $\mathrm{a}=17 \mathrm{~b}$ and constant term $\left.=-3 b^{2}\right)$

Now, we will split the coefficient of a into two parts such that their sum is $17 \mathrm{~b}$ and their product equals the product of the coefficient of $\mathrm{a}^{2}$ and the constant term, i.e., $6 \times\left(-3 \mathrm{~b}^{2}\right)=-18 \mathrm{~b}^{2}$.

Now,

$18 \mathrm{~b}+(-\mathrm{b})=17 \mathrm{~b}$

and

$18 b \times(-b)=-18 b^{2}$

Replacing the middle term $17 \mathrm{ab}$ by $-\mathrm{ab}+18 \mathrm{ab}$, we get:

$16 a^{2}+17 a b-3 b^{2}=6 a^{2}+-a b+18 a b-3 b^{2}$

$=\left(6 a^{2}-a b\right)+\left(18 a b-3 b^{2}\right)$

$=a(6 a-b)+3 b(6 a-b)$

$=(a+3 b)(6 a-b)$