Question

Expand:
(i) (a + 2b + 5c)²
(ii) (2b − b + c)²
(iii) (a − 2b − 3c)²

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right){\left(a+2b+5c\right)}^2={\left(a\right)}^2+{\left(2b\right)}^2+{\left(5c\right)}^2+2\left(a\right)\left(2b\right)+2\left(2b\right)\left(5c\right)+2\left(5c\right)\left(a\right) \\ =a^2+4b^2+25c^2+4ab+20bc+10ac \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right){\left(2a-b+c\right)}^2={\left[\left(2a\right)+\left(-b\right)+\left(c\right)\right]}^2 \\ ={\left(2a\right)}^2+{\left(-b\right)}^2+{\left(c\right)}^2+2\left(2a\right)\left(-b\right)+2\left(-b\right)\left(c\right)+4\left(a\right)\left(c\right) \\ =4a^2+b^2+c^2-4ab-2bc+4ac \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right){\left(a-2b-3c\right)}^2={\left[a+\left(-2b\right)+\left(-3c\right)\right]}^2 \\ ={\left(a\right)}^2+{\left(-2b\right)}^2+{\left(-3c\right)}^2+2\left(a\right)\left(-2b\right)+2\left(-2b\right)\left(-3c\right)+2\left(a\right)\left(-3c\right) \\ =a^2+4b^2+9c^2-4ab+12bc-6ac \end{gathered}$$