Question

Factorise (1-x^2) (1-y^2) + 4xy

Answer 1

= (1-x^2)×(1-y^2)+4xy
= 1×(1-y^2)-x^2(1-y^2)+4xy
= 1-y^2-x^2+x^2 y^2 + 4xy
= (1+2xy+ x^2 y^2) - (y^2 + x^2 - 2xy)
= (1+xy)^2 - (x-y)^2
= [(1+xy)+(x-y)] × [( 1+xy)- (x-y)]
= [(1+x-y + xy) ×( 1-x+y + xy)]

Answer 2

$\left(1-x^{2}\right)\left(1-y^{2}\right)+4 x y=(1+x y+x-y)(1+x y-x+y)$

To find:

Factorize $\left(1-x^{2}\right)\left(1-y^{2}\right)+4 x y$

Solution:

$\left(1-x^{2}\right)\left(1-y^{2}\right)+4 x y$

$\begin{array}{l}{=1-y^{2}-x^{2}+x^{2} y^{2}+4 x y} \\ \\ {=1+x^{2} y^{2}-x^{2}-y^{2}+4 x y}\end{array}$

By splitting 4xy,

$=1+x^{2} y^{2}-x^{2}-y^{2}+2 x y+2 x y$

By regrouping terms,

$\begin{array}{l}{=1+x^{2} y^{2}+2 x y-x^{2}-y^{2}+2 x y} \\ \\ {=\left(1+x^{2} y^{2}+2 x y\right)-\left(x^{2}+y^{2}-2 x y\right)} \\ \\ {=\left[(1)^{2}+(x y)^{2}+2 \times 1 \times x y\right]-(x-y)^{2}}\end{array}$

Using identity:

$\left(a^{2}+b^{2}+2 a b\right)=(a+b)^{2}$

$=(1+x y)^{2}-(x-y)^{2}$

Using identity:

$\left(a^{2}-b^{2}\right)=(a+b)(a-b)$

$\bold{=(1+x y+x-y)(1+x y-x+y)}$.