Question

Factorise: ${(l+m)}^2-4lm$

Answer 1

We have

$(l+m{)}^2-4lm=l^2+m^2+2lm-4lm$

$=l^2+m^2-2ml$

$=l^2-2\left(l\right)\left(m\right)+m^2$

$=(l-m{)}^2$

$=(l-m)(l-m)$

$\therefore (l+m{)}^2-4lm=(l-m\left)\right(l-m)$