Question

Question 83 (xix)

Multiply the following :

$\frac{3}{2}{p}^2+\frac{2}{3}{q}^2,(2p^2-3q^2)$

Answer 1

We have, $(\frac{3}{2}{p}^2+\frac{2}{3}{q}^2)and(2p^2-3q^2)$
$\therefore (\frac{3}{2}{p}^2+\frac{2}{3}{q}^2)(2p^2-3q^2)=\frac{3}{2}{p}^2(2p^2-3q^2)+\frac{2}{3}{q}^2(2p^2-3q^2)$
$=\frac{3}{2}{p}^2\times 2p^2-\frac{9}{2}{p}^2{q}^2+\frac{4}{3}{q}^2{p}^2-2q^4$
$=3p^4+(\frac{4}{3}-\frac{9}{2})p^2{q}^2-2q^4$
$=3p^4+\left(\frac{8-27}{6}\right)p^2{q}^2-2q^4$
$=3p^4-\frac{19}{6}{p}^2{q}^2-2q^4$