Question
Multiply the following:$\frac{3}{2}\text{p}^2+\frac{2}{3}\text{q}^2,\big(2\text{p}^2-3\text{q}^2\big)$

Answer

$\frac{3}{2}\text{p}^2+\frac{2}{3}\text{q}^2,\big(2\text{p}^2-3\text{q}^2\big)$$\Big(\frac{3}{2}\text{p}^2+\frac{2}{3}\text{q}^2\Big)\big(2\text{p}^2-3\text{q}^2\big)\\=\frac{3}{2}\text{p}^2\big(2\text{p}^2-3\text{q}^2\big)+\frac{2}{3}\text{q}^2\big(2\text{p}^2-3\text{q}^2\big)$
$=\frac{3}{2}\text{p}^2\times2\text{p}^2-\frac{9}{2}\text{p}^2\text{q}^2+\frac{4}{3}\text{q}^2\text{p}^2-2\text{q}^4$
$=3\text{p}^4+\Big(\frac{4}{3}-\frac{9}{2}\Big)\text{p}^2\text{q}^2-2\text{q}^4$
$=3\text{p}^4+\Big(\frac{8-27}{6}\Big)\text{p}^2\text{q}^2-2\text{q}^4$
$=3\text{p}^4-\frac{19}{6}\text{p}^2\text{q}^2-2\text{q}^4$

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