Question

Factorise the following trinomials:
$\frac{x^2}{4y^2}-\frac{2}{3}+\frac{4y^2}{9x^2}$

Answer 1

$\frac{x^2}{4y^2}-\frac{2}{3}+\frac{4y^2}{9x^2}$

$=\frac{1}{4}\left(\frac{x^2}{y^2}\right)-\frac{2}{3}+\frac{4}{9}\left(\frac{y^2}{x^2}\right)$

Substituting $\frac{x^2}{y^2}=t$, we have

$=\frac{1}{4}\left(t\right)-\frac{2}{3}+\frac{4}{9}\left(\frac{1}{t}\right)$

$=\frac{(\frac{1}{4}{t}^2-\frac{2}{3}t+\frac{4}{9})}{t}$

$=\frac{(\frac{1}{4}{t}^2-\frac{1}{3}t-\frac{1}{3}t+\frac{4}{9})}{t}$

$=\frac{(\frac{1}{4}{t}^2-\left(\frac{4}{4}\right)\frac{1}{3}t-\frac{1}{3}t+\frac{4}{9})}{t}$

$=\frac{(\frac{1}{4}t(t-\frac{4}{3})-\frac{1}{3}(t-\frac{4}{3}))}{t}$

$=\frac{(t-\frac{4}{3})(\frac{1}{4}t-\frac{1}{3})}{t}$

As $t=\frac{x^2}{y^2}$

$=\frac{(\frac{x^2}{y^2}-\frac{4}{3})(\frac{1}{4}\frac{x^2}{y^2}-\frac{1}{3})}{\frac{x^2}{y^2}}$

$=\left(\frac{y^2}{x^2}\right)\times (\frac{x^2}{y^2}-\frac{4}{3})(\frac{1}{4}\frac{x^2}{y^2}-\frac{1}{3})$

$(1-\frac{4}{3}\left(\frac{y^2}{x^2}\right))(\frac{1}{4}\frac{x^2}{y^2}-\frac{1}{3})$

Hence the factors of $\frac{x^2}{4y^2}-\frac{2}{3}+\frac{4y^2}{9x^2}$ are $(1-\frac{4}{3}\left(\frac{y^2}{x^2}\right))(\frac{1}{4}\frac{x^2}{y^2}-\frac{1}{3})$.