Question

Find the square of the following:
$\left(iv\right)(2\text{a}+3\text{b}-4\text{c})$

Answer 1

To find: $(2\text{a}+3\text{b}-4\text{c}{)}^2$

Now, expand the given expression by using the square of sum identity

$(\text{a}+\text{b}+\text{c}{)}^2={\text{a}}^2+{\text{b}}^2+{\text{c}}^2+2\text{a}\text{b}+2\text{b}\text{c}+2\text{c}\text{a}$.

$\Rightarrow (2\text{a}+3\text{b}-4\text{c}{)}^2=(2\text{a}{)}^2+(3\text{b}{)}^2+(-4\text{c}{)}^2+2\times 2\text{a}\times 3\text{b}+2\times 3\text{b}\times (-4\text{c})+2\times 2\text{a}\times (-4\text{c})$

$=4{\text{a}}^2+9{\text{b}}^2+16{\text{c}}^2+12\text{a}\text{b}-24\text{b}\text{c}-16\text{a}\text{c}$

Hence, square of $(2\text{a}+3\text{b}-4\text{c})$ is $4{\text{a}}^2+9{\text{b}}^2+16{\text{c}}^2+12\text{a}\text{b}-24\text{b}\text{c}-16\text{a}\text{c}$