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) ${\left(\frac{3x}{4}+\frac{2y}{9}\right)}^2$
(vi) (9x − 10)²
(vii) (x²y − yz²)²
(viii) ${\left(\frac{x}{y}-\frac{y}{x}\right)}^2$
(ix) ${\left(3m-\frac{4}{5}n\right)}^2$

Answer 1

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

(i) We have:
$$\begin{gathered} (8a+3b{)}^2 \\ ={\left(8a\right)}^2+2\times 8a\times 3b+{\left(3b\right)}^2 \\ =64a^2+48ab+9b^2 \end{gathered}$$

(ii)We have:
$$\begin{gathered} (7x+2y{)}^2 \\ ={\left(7x\right)}^2+2\times 7x\times 2y+{\left(2y\right)}^2 \\ =49x^2+28xy+4y^2 \end{gathered}$$

(iii) We have :
$$\begin{gathered} (5x+11{)}^2 \\ ={\left(5x\right)}^2+2\times 5x\times 11+{\left(11\right)}^2 \\ =25x^2+110x+121 \end{gathered}$$

(iv) We have:
$$\begin{gathered} {\left({\displaystyle \frac{a}{2}}+{\displaystyle \frac{2}{a}}\right)}^2 \\ ={\left(\frac{a}{2}\right)}^2+2\times \frac{a}{2}\times \frac{2}{a}+{\left(\frac{2}{a}\right)}^2 \\ ={\frac{a}{4}}^2+2+\frac{4}{a^2} \end{gathered}$$

(v) We have:
$$\begin{gathered} {\left({\displaystyle \frac{3x}{4}}+{\displaystyle \frac{2y}{9}}\right)}^2 \\ ={\left(\frac{3x}{4}\right)}^2+2\times \frac{3x}{4}\times \frac{2y}{9}+{\left(\frac{2y}{9}\right)}^2 \\ ={\frac{9x}{16}}^2+\frac{1}{3}xy+\frac{4y^2}{81} \end{gathered}$$

(vi) We have:
$$\begin{gathered} (9x-10{)}^2 \\ {\left(9x\right)}^2-2\times 9x\times 10+{\left(10\right)}^2 \\ =81x^2-180x+100 \end{gathered}$$

(vii) We have:
$$\begin{gathered} (x^{2}y-y{z}^2{)}^2 \\ {\left(x^{2}y\right)}^2-2\times x^{2}y\times y{z}^2+{\left(y{z}^2\right)}^2 \\ =x^4{y}^2-2x^2{y}^2{z}^2+y^2{z}^4 \end{gathered}$$

(viii) We have:
$$\begin{gathered} {\left({\displaystyle \frac{x}{y}}-{\displaystyle \frac{y}{x}}\right)}^2 \\ ={\left(\frac{x}{y}\right)}^2-2\times \frac{x}{y}\times \frac{y}{x}+{\left(\frac{y}{x}\right)}^2 \\ =\frac{x^2}{y^2}-2+\frac{y^2}{x^2} \end{gathered}$$

(ix) We have:
$$\begin{gathered} {\left(3m-{\displaystyle \frac{4}{5}}n\right)}^2 \\ ={\left(3m\right)}^2-2\times 3m\times \frac{4}{5}n+{\left(\frac{4}{5}n\right)}^2 \\ =9m^2-\frac{24mn}{5}+\frac{16}{25}{n}^2 \end{gathered}$$