Question

Factorise by using formula $4{\left( {2a + b} \right)^2} - {\left( {a - b} \right)^2}$
$
  A.3\left( {a + b} \right)\left( {a + b} \right) \\
  B.3\left( {5a + b} \right)\left( b \right) \\
  C.3\left( {5a + b} \right)\left( {a + b} \right) \\
  D.3\left( {5a - b} \right)\left( {a + b} \right) \\
 $

Answer 1

Hint – Here we will proceed by factorisation of quadratic polynomials of the form ${x^2} + bx + c$., we have to find numbers p and q such that p + q = b and p$ \times $q = c.

Complete step-by-step answer:
It is given that –
$4{\left( {2a + b} \right)^2} - {\left( {a - b} \right)^2}$
By using the identity ${\left( {a + b} \right)^2}$ and ${\left( {a - b} \right)^2}$-
$ \Rightarrow $ 4$\left( {4{a^2} + 4ab + {b^2}} \right) - \left( {{a^2} - 2ab + {b^2}} \right) = 0$
 $ \Rightarrow 16{a^2} + 16ab + 4{b^2} - {a^2} + 2ab - {b^2} = 0$
 By simplifying we will get –
$ \Rightarrow 15{a^2} + 18ab + 3{b^2} = 0$
By taking 3 as common we will get,
$ = 3\left( {5{a^2} + 6ab + {b^2}} \right)$
By splitting the middle term, we will get-
$ = 3\left( {5{a^2} + 5ab + ab + {b^2}} \right)$
By taking 5a and b as common we will get –
$
   = 3\left[ {5a\left( {a + b} \right) + b\left( {a + b} \right)} \right] \\
   = 3\left( {5a + b} \right)\left( {a + b} \right) \\
 $
Therefore, the right answer is option C.
Note- Whenever we come up with this type of problem, one must know that there are various methods to factorise an equation that is the highest common factor. Factorization using the common factor, Quadratic trinomials, use of perfect squares. (Here we used splitting the middle term).