Question

Expand:
(i) (2a − 5b − 7c)²
(ii) (−3a + 4b − 5c)²
(iii) ${\left(\frac{1}{2}a-\frac{1}{4}a+2\right)}^2$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right){\left(2a-5b-7c\right)}^2={\left[\left(2a\right)+\left(-5b\right)+\left(-7c\right)\right]}^2 \\ ={\left(2a\right)}^2+{\left(-5b\right)}^2+{\left(-7c\right)}^2+2\left(2a\right)\left(-5b\right)+2\left(-5b\right)\left(-7c\right)+2\left(2a\right)\left(-7c\right) \\ =4a^2+25b^2+49c^2-20ab+70bc-28ac \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right){\left(-3a+4b-5c\right)}^2={\left[\left(-3a\right)+\left(4b\right)+\left(-5c\right)\right]}^2 \\ ={\left(-3a\right)}^2+{\left(4b\right)}^2+{\left(-5c\right)}^2+2\left(-3a\right)\left(4b\right)+2\left(4b\right)\left(-5c\right)+2\left(-3a\right)\left(-5c\right) \\ =9a^2+16b^2+25c^2-24ab-40bc+30ac \end{gathered}$$

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