Factorize each of the following algebraic expression:

Solution:

$\left(a^{2}-5 a\right)^{2}-36$

$=\left(a^{2}-5 a\right)^{2}-6^{2}$

$=\left[\left(a^{2}-5 a\right)-6\right]\left[\left(a^{2}-5 a\right)+6\right]$

$=\left(a^{2}-5 a-6\right)\left(a^{2}-5 a+6\right)$

In order to factorise $a^{2}-5 a-6$, we will find two numbers $p$ and $q$ such that $p+q=-5$ and $p q=-6$

Now,

$(-6)+1=-5$

and

$(-6) \times 1=-6$

Splitting the middle term $-5$ in the given quadratic as $-6 a+a$, we get $:$

$a^{2}-5 a-6=a^{2}-6 a+a-6$

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

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

$=(a+1)(a-6)$

Now,

In order to factorise $a^{2}-5 a+6$, we will find two numbers $p$ and $q$ such that $p+q=-5$ and $p q=6$

Clearly,

$(-2)+(-3)=-5$

and

$(-2) \times(-3)=6$

Splitting the middle term $-5$ in the given quadratic as $-2 a-3 a$, we get:

$a^{2}-5 a+6=a^{2}-2 a-3 a+6$

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

$=a(a-2)-3(a-2)$

$=(a-3)(a-2)$

$\therefore\left(a^{2}-5 a-6\right)\left(a^{2}-5 a+6\right)=(a-6)(a+1)(a-3)(a-2)$

$=(a+1)(a-2)(a-3)(a-6)$