Question:
Factorise:
m2 − 4mn + 4n2
Solution:
We have:
$m^{2}-4 m n+4 n^{2}=m^{2}-2 \times m \times 2 n+(2 n)^{2}$
$=(m-2 n)^{2}$
$\therefore m^{2}-4 m n+4 n^{2}=(m-2 n)^{2}$
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