Question

Expand ${[\frac{1}{4}a-\frac{1}{2}b+1]}^2$

Answer 1

${[\frac{1}{4}a-\frac{1}{2}b+1]}^2\left[Usingtheidentity\right(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

$\Rightarrow {\left(\frac{1}{4}a\right)}^2+{\left(\frac{-1}{2}b\right)}^2+(1{)}^2+2\left(\frac{1}{4}a\right)\left(\frac{-1}{2}b\right)+2\left(\frac{-1}{2}b\right)\times 1+2(1)\times \frac{1}{4}a=\frac{1}{16}{a}^2+\frac{1}{4}{b}^2+1-\frac{1}{4}ab-b+\frac{1}{2}a=\frac{a^2}{16}+\frac{b^2}{4}+1-\frac{1}{4}ab-b+\frac{1}{2}a=\frac{a^2}{16}+\frac{b^2}{4}+1-\frac{ab}{4}-b+\frac{a}{2}$