Question

Expand ${\left(\frac{1}{2}a-\frac{1}{4}b+2\right)}^2$.

Answer 1

Expanding the given expression:

The given expression is ${\left(\frac{1}{2}a-\frac{1}{4}b+2\right)}^2$.

Use identity, $(x+y+z{)}^2=x^2+y^2+z^2+2xy+2yz+2xz$ to expand.

The given expression can be rewritten as ${\left(\frac{1}{2}a-\frac{1}{4}b+2\right)}^2={\left[\frac{1}{2}a+\left(-\frac{1}{4}b\right)+2\right]}^2$.

Comparing expression with the identity we get, $x=\frac{1}{2}a,y=-\frac{1}{4}b,z=2$.

Substitute values in identity we get:

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

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