Question

Factorize each of the following algebraic expression:
(a² − 5a)² − 36

Answer 1

\[( a^2 - 5a )^2 - 36\]
\[ = ( a^2 - 5a )^2 - 6^2 \]
\[ = [( a^2 - 5a) - 6][( a^2 - 5a) + 6]\]
\[ = ( a^2 - 5a - 6)( a^2 - 5a + 6)\]
\[\text{ In order to factorise }a^2 - 5a - 6, \text{ we will find two numbers p and q such that }p + q = - 5\text{ and } pq = - 6\]
Now, 
\[( - 6) + 1 = - 5 \]
and 
\[( - 6) \times 1 = - 6\]
\[\text{ Splitting the middle term } - 5\text{ in the given quadratic as }- 6a + a, \text{ we get: }\]
\[ a^2 - 5a - 6 = a^2 - 6a + a - 6\]
\[ = ( a^2 - 6a) + (a - 6)\]
\[ = a(a - 6) + (a - 6)\]
\[ = (a + 1)(a - 6)\]
Now, 
\[\text{ In order to factorise }a^2 - 5a + 6, \text{ we will find two numbers p and q such that }p + q = - 5\text{ and } pq = 6\]
Clearly, 
\[( - 2) + ( - 3) = - 5 \]
and 
\[( - 2) \times ( - 3) = 6\]
\[\text{ Splitting the middle term }- 5\text{ in the given quadratic as }- 2a - 3a,\text{ we get: }\]
\[ a^2 - 5a + 6 = a^2 - 2a - 3a + 6\]
\[ = ( a^2 - 2a) - (3a - 6)\]
\[ = a(a - 2) - 3(a - 2)\]
\[ = (a - 3)(a - 2)\]
\[ \therefore ( a^2 - 5a - 6)( a^2 - 5a + 6) = (a - 6)(a + 1)(a - 3)(a - 2)\]
\[ = (a + 1)(a - 2)(a - 3)(a - 6)\]