Factorize each of the following algebraic expression:

(a + 7)(a − 10) + 16

$=a^{2}-10 a+7 a-70+16$

$=a^{2}-3 a-54$

To factorise $a^{2}-3 a-54$, we will find two numbers $p$ and $q$ such that $p+q=-3$ and $p q=-54$.

Now,

$6+(-9)=-3$

and

$6 \times(-9)=-54$

Splitting the middle term $-3 a$ in the given quadratic as $-9 a+6 a$, we get :

$a^{2}-3 a-54=a^{2}-9 a+6 a-54$

$=\left(a^{2}-9 a\right)+(6 a-54)$

$=a(a-9)+6(a-9)$

$=(a+6)(a-9)$