Question:
Factorise:
(l + m)2 − 4lm
Solution:
We have:
$(l+m)^{2}-4 l m=\left(l^{2}+m^{2}+2 l m\right)-4 l m$
$=l^{2}+m^{2}+2 l m-4 l m$
$=l^{2}+m^{2}-2 l m$
$=(l)^{2}+(m)^{2}-2 \times l \times m$
$=(l-m)^{2}$
$\therefore(l+m)^{2}-4 l m=(l-m)^{2}$
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