Resolve each of the following quadratic trinomial into factor:

The given expression is $36 \mathrm{a}^{2}+12 \mathrm{abc}-15 \mathrm{~b}^{2} \mathrm{c}^{2} .       \quad$ (Coefficient of $\mathrm{a}^{2}=36$, coefficient of $\mathrm{a}=12 \mathrm{bc}$ and constant term $=-15 \mathrm{~b}^{2} \mathrm{c}^{2}$ )

Now, we will split the coefficient of a into two parts such that their sum is 12 bc and their product equals the product of the coefficient of $\mathrm{a}^{2}$ and the constant term, i.e., $36 \times\left(-15 \mathrm{~b}^{2} \mathrm{c}^{2}\right)=-540 \mathrm{~b}^{2} \mathrm{c}^{2}$.

Now,

$(-18 \mathrm{bc})+30 \mathrm{bc}=12 \mathrm{bc}$

and

$(-18 \mathrm{bc}) \times 30 \mathrm{bc}=-540 \mathrm{~b}^{2} \mathrm{c}^{2}$

Replacing the middle term 12 abc by $-18 a b c+30 a b c$, we get:

$36 \mathrm{a}^{2}+12 \mathrm{abc}-15 \mathrm{~b}^{2} \mathrm{c}^{2}=36 \mathrm{a}^{2}-18 \mathrm{abc}+30 \mathrm{abc}-15 \mathrm{~b}^{2} \mathrm{c}^{2}$

$=\left(36 \mathrm{a}^{2}-18 \mathrm{abc}\right)+\left(30 \mathrm{abc}-15 \mathrm{~b}^{2} \mathrm{c}^{2}\right)$

$=18 \mathrm{a}(2 \mathrm{a}-\mathrm{bc})+15 \mathrm{bc}(2 \mathrm{a}-\mathrm{bc})$

$=(18 \mathrm{a}+15 \mathrm{bc})(2 \mathrm{a}-\mathrm{bc})$

$=3(6 \mathrm{a}+5 \mathrm{bc})(2 \mathrm{a}-\mathrm{bc})$