Question Number 173226 by AgniMath last updated on 08/Jul/22
$$\:\:\:\:\boldsymbol{\mathrm{Factorize}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\:\left({a}\:+\:{b}\right)^{\mathrm{2}} \:−\:\frac{\mathrm{9}}{\mathrm{16}}\:\left(\mathrm{2}{a}\:−\:{b}\right)^{\mathrm{2}} \\ $$
Commented by mr W last updated on 08/Jul/22
![=(((a+b)/2))^2 −[((3(2a−b))/4)]^2 =(((a+b)/2)+((3(2a−b))/4))(((a+b)/2)−((3(2a−b))/4)) =(((8a−b)(5b−4a))/(16))](https://www.tinkutara.com/question/Q173239.png)
$$=\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} −\left[\frac{\mathrm{3}\left(\mathrm{2}{a}−{b}\right)}{\mathrm{4}}\right]^{\mathrm{2}} \\ $$$$=\left(\frac{{a}+{b}}{\mathrm{2}}+\frac{\mathrm{3}\left(\mathrm{2}{a}−{b}\right)}{\mathrm{4}}\right)\left(\frac{{a}+{b}}{\mathrm{2}}−\frac{\mathrm{3}\left(\mathrm{2}{a}−{b}\right)}{\mathrm{4}}\right) \\ $$$$=\frac{\left(\mathrm{8}{a}−{b}\right)\left(\mathrm{5}{b}−\mathrm{4}{a}\right)}{\mathrm{16}} \\ $$
Commented by peter frank last updated on 08/Jul/22
$$\mathrm{thanks} \\ $$