Question

State True or False.

${[\frac{1}{4}a-\frac{1}{2}b+1]}^2$$=\frac{1}{16}{a}^2+\frac{1}{4}{b}^2+1-\frac{1}{4}ab-b+\frac{a}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

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