Question

Expand.
(i) (2a – 3b)²

(ii) (10 + y)²

(iii) ${\left(\frac{p}{3}+\frac{q}{4}\right)}^2$

(iv) ${\left(y-\frac{3}{y}\right)}^2$

Answer 1

(i) (2a – 3b)²
Using the identity ${\left(x-y\right)}^2=x^2-2xy+y^2$
${\left(2a-3b\right)}^2={\left(2a\right)}^2-2\times 2a\times 3b+{\left(3b\right)}^2=4a^2-12ab+9b^2$

(ii) (10 + y)²
Using the identity: ${\left(a+b\right)}^2=a^2+2ab+b^2$
$$\begin{gathered} ={\left(10\right)}^2+2\times 10\times y+y^2 \\ =100+20y+y^2 \end{gathered}$$

(iii) ${\left(\frac{p}{3}+\frac{q}{4}\right)}^2$
Using the identity: ${\left(a+b\right)}^2=a^2+2ab+b^2$
$$\begin{gathered} ={\left(\frac{p}{3}\right)}^2+2\times \frac{p}{3}\times \frac{q}{4}+{\left(\frac{q}{4}\right)}^2 \\ =\frac{p^2}{9}+\frac{pq}{6}+\frac{q^2}{16} \end{gathered}$$

(iv) ${\left(y-\frac{3}{y}\right)}^2$
Using the identity ${\left(a-b\right)}^2=a^2-2ab+b^2$
$$\begin{gathered} =y^2-2\times y\times \frac{3}{y}+{\left(\frac{3}{y}\right)}^2 \\ =y^2-6+\frac{9}{y^2} \end{gathered}$$