Question

select a suitable identity and find the following products
$(3k+4l)(3k+4l)$

Answer 1

$(3k+4l)(3k+4l)$

$={(3k+4l)}^2$

$={\left(3k\right)}^2+{\left(4l\right)}^2+2\left(3k\right)\left(4l\right)[\text{Using identity:}{(a+b)}^2=a^2+b^2+2ab]$

$=9k^2+16l^2+24kl$