# What is the standard form of  f=(x - 2)(x - 2)(x + y)(x - y) ?

Dec 11, 2017

To find the standard form of $f$, we need to first expand the brackets and rearrange them in a descending power of degree.

$f = \left(x - 2\right) \left(x - 2\right) \left(x + y\right) \left(x - y\right)$
$= {\left(x - 2\right)}^{2} \cdot \left(x + y\right) \left(x - y\right)$

we can use identities to expand it.
Identities :

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

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

$f = \left({x}^{2} - 2 \left(x\right) \left(2\right) + {2}^{2}\right) \left({x}^{2} - {y}^{2}\right)$
$= \left({x}^{2} - 4 x + 4\right) \left({x}^{2} - {y}^{2}\right)$
$= \left({x}^{2}\right) \left({x}^{2} - {y}^{2}\right) - 4 x \left({x}^{2} - {y}^{2}\right) + 4 \left({x}^{2} - {y}^{2}\right)$
$= {x}^{4} - {x}^{2} {y}^{2} - 4 {x}^{3} + 4 x {y}^{2} + 4 {x}^{2} - 4 {y}^{2}$

Remarks: ${x}^{2} {y}^{2}$ have a degree of $4$, where $2$ from ${x}^{2}$ and $2$ from ${y}^{2}$

As it is already in a descending degree of power, we don't have to rearrange it and it's the answer. Hope this can help you.