(i)

πn²k - $\frac{1}{4}$πn²Q

πn² is common so factorize them out.

πn²k - $\frac{1}{4}$πn²Q = πn²(k - $\frac{1}{4}$Q)

πn²k - $\frac{1}{4}$πn²Q = πn²(k - $\frac{\mathrm{1\; x\; Q}}{4}$)

πn²k - $\frac{1}{4}$πn²Q = πn²(k - $\frac{\mathrm{Q}}{4}$)

(ii)

π = $\frac{22}{7}$, n = 2, k = 19 and Q = 20

πn²(k - $\frac{\mathrm{Q}}{4}$) = $\frac{22}{7}$ x 2²(19 - $\frac{20}{4}$)

πn²(k - $\frac{\mathrm{Q}}{4}$) = $\frac{22}{7}$ x 4(19 - 5)

Note: 2² = 2 x 2 = 4
4 divides itself once and 20, 5 times

πn²(k - $\frac{\mathrm{Q}}{4}$) = $\frac{22}{7}$ x 4(14)

πn²(k - $\frac{\mathrm{Q}}{4}$) = $\frac{22}{7}$ x 4 x 14

Note: 7 divides itself once and 14, 2 times.

πn²(k - $\frac{\mathrm{Q}}{4}$) = 22 x 4 x 2

πn²(k - $\frac{\mathrm{Q}}{4}$) = 176