Factorise:20a2 − 45b2
We have:
$20 a^{2}-45 b^{2}=5\left(4 a^{2}-9 b^{2}\right)$
$=5\left\{(2 a)^{2}-(3 b)^{2}\right\}$
$=5(2 a+3 b)(2 a-3 b)$
$\therefore 20 a^{2}-45 b^{2}=5(2 a+3 b)(2 a-3 b)$
Leave a comment
All Study Material