Question

Factorise: $4xy-x^2-4y^2+z^2$

Answer 1

Since, we have $\left[\right(a^2-b^2)=(a-b\left)\right(a+b\left)\right]$ and
$\left[\right(a-b{)}^2=a^2-2ab+b^2]$

We will use above two identities to simplify given expression

$4xy-x^2-4y^2+z^2=-x^2+4xy-4y^2+z^2=-(x^2-4xy+4y^2)+z^2=-[x^2-2\times x\times 2y+(2y{)}^2]+z^2=-(x-2y{)}^2+z^2[\because (a-b{)}^2=a^2-2ab+b^2]=z^2-(x-2y{)}^2=[z+(x-2y\left)\right][z-(x-2y\left)\right]=[z+x-2y][z-x+2y]$

Therefore, $4xy-x^2-4y^2+z^2=[z+x-2y][z-x+2y]$