Factoring by grouping: $x^4 - y^4 -4x^2 + 4$

Question

Please help me factor $x^4 - y^4 -4x^2 + 4$ by grouping terms. Thank you.

Answer 1

Group all the $x$ terms together and all the $y$ terms together: $$(x^4-4x^2+4)-y^4=(x^2-2)^2-(y^2)^2=(x^2-y^2-2)(x^2+y^2-2)$$ using $a^2-b^2=(a-b)(a+b)$.

Answer 2

$$x^4-y^4-4x^2+4=(x^4-4x^2+4)-y^4=(x^2-2)^2-(y^2)^2=(x^2-y^2-2)(x^2+y^2-2)$$

Answer 3

$x^4 - y^4 -4x^2 + 4 $

$= (x^4 -4x^2 + 4) - y^4$

$= [(x^2)^2 -2(2x^2) + 2^2] - (y^2)^2$

The expression $(x^2)^2 -2(2x^2) + 2^2$ has the form: $a^2 -2ab + b^2 = (a - b)^2$

where $a=x^2$ and $b=2$

Thus, $(x^2)^2 -2(2x^2) + 2^2 = (x^2 - 2)^2$

Hence

$[(x^2)^2 -2(2x^2) + 2^2] - (y^2)^2$

$= (x^2 - 2)^2 -(y^2)^2$

which is a difference of two squares i.e. has the form: $a^2-b^2 = (a+b)(a-b)$

where $a= x^2-2$ and $ b = y^2$

Thus, $(x^2 - 2)^2 -(y^2)^2 = (x^2 -2 + y^2)(x^2 -2 - y^2)$

$\therefore $ $x^4 - y^4 -4x^2 + 4 = (x^2 -2 + y^2)(x^2 -2 - y^2)$ Answer