Question

Factorise the following expression $81{(2a + b)^2} - 25{(a - 2b)^2}$

Answer 1

Hint:In the above expression we have two terms having minus sign between them. And we can clearly see both terms are perfect squares, so we use algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$ and factorise the given expression..

Complete step-by-step answer:
We are given the expression,
$81{(2a + b)^2} - 25{(a - 2b)^2}$ $ \to (1)$
In order to factorize the above expression, we will try to make its factors or we will try to get something common.
Consider $81{(2a + b)^2} - 25{(a - 2b)^2}$
$ = {\left[ {9(2a + b)} \right]^2} - {\left[ {5(a - 2b)} \right]^2}$
Now using identity ${a^2} - {b^2} = (a + b)(a - b)$, we get
$
   = \left[ {9(2a + b) - \left[ {5(a - 2b)} \right]} \right] \left[ {9(2a + b) + 5(a - 2b)} \right] \\
   = \left[ {18a + 9b - 5a + 10b} \right] \left[ {18a + 9b + 5a - 10b} \right] \\
   = (13a + 19b)(23a - b) \\
 $
Therefore $81{(2a + b)^2} - 25{(a - 2b)^2}$$ = (13a + 19b)(23a - b$$)$
So, the factors of (1) are $(13a + 19b)$ and $(23a - b)$

Note:In these types of questions, we need to observe the suitable algebraic identity which can be used. There are few important algebraic identities-
$
  {(a \pm b)^2} = {a^2} \pm 2ab + {b^2} \\
  {a^2} - {b^2} = (a - b)(a + b) \\
  {(a \pm b)^3} = {a^3} \pm {b^3} + 3ab(a \pm b) \\ $
Now in the above question we are given $81{(2a + b)^2} - 25{(a - 2b)^2}$, by observing expression carefully one can easily find that ${a^2} - {b^2} = (a - b)(a + b)$ will be used here. So only by keeping the algebraic identities in mind one can easily solve this type of questions.