# Expand:

Question:

Expand:

(i) (8a + 3b)2

(ii) (7x + 2y)2

(iii) (5x + 11)2

(iv) $\left(\frac{a}{2}+\frac{2}{a}\right)^{2}$

(v) $\left(\frac{3 x}{4}+\frac{2 y}{9}\right)^{2}$

(vi) (9x − 10)2

(vii) (x2y − yz2)2

(viii) $\left(\frac{x}{y}-\frac{y}{x}\right)^{2}$

(ix) $\left(3 m-\frac{4}{5} n\right)^{2}$

Solution:

We shall use the identities (a+b)2 =a2 +b2 +2ab and (a-b)2 =a2 +b2 -2ab.

(i) We have:

$(8 a+3 b)^{2}$

$=(8 a)^{2}+2 \times 8 a \times 3 b+(3 b)^{2}$

$=64 a^{2}+48 a b+9 b^{2}$

(ii) We have:

$(7 x+2 y)^{2}$

$=(7 x)^{2}+2 \times 7 x \times 2 y+(2 y)^{2}$

$=49 x^{2}+28 x y+4 y^{2}$

(iii) We have:

$(5 x+11)^{2}$

$=(5 x)^{2}+2 \times 5 x \times 11+(11)^{2}$

$=25 x^{2}+110 x+121$

(iv)We have:

$\left(\frac{a}{2}+\frac{2}{a}\right)^{2}$

$=\left(\frac{a}{2}\right)^{2}+2 \times \frac{a}{2} \times \frac{2}{a}+\left(\frac{2}{a}\right)^{2}$

$=\frac{a}{4}^{2}+2+\frac{4}{a^{2}}$

(v) We have:

$\left(\frac{3 x}{4}+\frac{2 y}{9}\right)^{2}$

$=\left(\frac{3 x}{4}\right)^{2}+2 \times \frac{3 x}{4} \times \frac{2 y}{9}+\left(\frac{2 y}{9}\right)^{2}$

$=\frac{9 x^{2}}{16}+\frac{1}{3} x y+\frac{4 y^{2}}{81}$

(vi) We have:

$(9 x-10)^{2}$

$(9 x)^{2}-2 \times 9 x \times 10+(10)^{2}$

$=81 x^{2}-180 x+100$

(vii) We have:

$\left(x^{2} y-y z^{2}\right)^{2}$

$\left(x^{2} y\right)^{2}-2 \times x^{2} y \times y z^{2}+\left(y z^{2}\right)^{2}$

$=x^{4} y^{2}-2 x^{2} y^{2} z^{2}+y^{2} z^{4}$

(viii)  We have:

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

$=\left(\frac{x}{y}\right)^{2}-2 \times \frac{x}{y} \times \frac{y}{x}+\left(\frac{y}{x}\right)^{2}$

$=\frac{x^{2}}{y^{2}}-2+\frac{y^{2}}{x^{2}}$

(ix)  We have:

$\left(3 m-\frac{4}{5} n\right)^{2}$

$=(3 m)^{2}-2 \times 3 m \times \frac{4}{5} n+\left(\frac{4}{5} n\right)^{2}$

$=9 m^{2}-\frac{24 m n}{5}+\frac{16}{25} n^{2}$