Question

Find each of the following products:
(i) (x − 4)(x − 4)
(ii) (2x − 3y)(2x − 3y)
(iii) $\left(\frac{3}{4}x-\frac{5}{6}y\right)\left(\frac{3}{4}x-\frac{5}{6}y\right)$
(iv) $\left(x-\frac{3}{x}\right)\left(x-\frac{3}{x}\right)$
(v) $\left(\frac{1}{3}{x}^2-9\right)\left(\frac{1}{3}{x}^2-9\right)$
(vi) $\left(\frac{1}{2}{y}^2-\frac{1}{3}y\right)\left(\frac{1}{2}{y}^2-\frac{1}{3}y\right)$

Answer 1

(i) We have:
$$\begin{gathered} (x-4)(x-4) \\ =(x-4{)}^2 \\ =x^2-2\times x\times 4+4^2[\mathrm{using}(a-b{)}^2=a^2-2ab+b^2] \\ =x^2-8x+16 \end{gathered}$$

(ii) We have:
$$\begin{gathered} (2x-3y)(2x-3y) \\ =(2x-3y{)}^2 \\ ={\left(2x\right)}^2-2\times 2x\times 3y+{\left(3y\right)}^2[\mathrm{using}(a-b{)}^2=a^2-2ab+b^2] \\ =4x^2-12xy+9y^2 \end{gathered}$$

(iii) We have:
$$\begin{gathered} \left(\frac{3}{4}x-\frac{5}{6}y\right)\left(\frac{3}{4}x-\frac{5}{6}y\right) \\ ={\left(\frac{3}{4}x-\frac{5}{6}y\right)}^2 \\ ={\left(\frac{3}{4}x\right)}^2-2\times \frac{3}{4}x\times \frac{5}{6}y+{\left(\frac{5}{6}y\right)}^2[\mathrm{using}(a-b{)}^2=a^2-2ab+b^2] \\ =\frac{9}{16}{x}^2-\frac{15}{12}xy+\frac{25}{36}{y}^2 \end{gathered}$$

(iv) We have:
$$\begin{gathered} \left(x-\frac{3}{x}\right)\left(x-\frac{3}{x}\right) \\ ={\left(x-\frac{3}{x}\right)}^2 \\ ={\left(x\right)}^2-2\times x\times \frac{3}{x}+{\left(\frac{3}{x}\right)}^2[\mathrm{using}(a-b{)}^2=a^2-2ab+b^2] \\ =x^2-6+\frac{9}{x^2} \end{gathered}$$

(v) We have:
$$\begin{gathered} \left(\frac{1}{3}{x}^2-9\right)\left(\frac{1}{3}{x}^2-9\right) \\ ={\left(\frac{1}{3}{x}^2-9\right)}^2 \\ ={\left(\frac{1}{3}{x}^2\right)}^2-2\times \frac{1}{3}{x}^2\times 9+{\left(9\right)}^2[\mathrm{using}(a-b{)}^2=a^2-2ab+b^2] \\ =\frac{1}{9}{x}^4-6x^2+81 \end{gathered}$$

(vi) We have:
$$\begin{gathered} \left(\frac{1}{2}{y}^2-\frac{1}{3}y\right)\left(\frac{1}{2}{y}^2-\frac{1}{3}y\right) \\ ={\left(\frac{1}{2}{y}^2-\frac{1}{3}y\right)}^2 \\ ={\left(\frac{1}{2}{y}^2\right)}^2-2\times \frac{1}{2}{y}^2\times \frac{1}{3}y+{\left(\frac{1}{3}y\right)}^2[\mathrm{using}(a-b{)}^2=a^2-2ab+b^2] \\ =\frac{1}{4}{y}^4-\frac{1}{3}{y}^3+\frac{1}{9}{y}^2 \end{gathered}$$