Question

How do you factor the trinomial `x^2y^2-5xy+4`?

Answer 1

`(xy-1)` `(xy-4)`

Explanation:

Break the expression into groups

(`x^2y^2-xy`) `+` `(-4xy+4)`

factor out common terms

`xy` `(xy-1)` `-4(xy-1)`

factor completely

`(xy-1)` `(xy-4)`

NOTE: the `xy-1` terms are listed twice when initially factoring out common terms. If you are factoring by grouping and you do not get one expression in parenthesis that is listed twice, you have done something wrong.

Answer 2

If the `x and y` together give you a problem think about it this way.

`(xy-1)(xy-4)`

Explanation:

Set `xy=a` giving:

`a^2-5a+4`

The whole number factors of 4 are `1xx4 and 2xx2`

Not that `4+1=5` but we need -5 so:

`(-1)xx(-4)=+4 and (-1)+(-4)=-5`

So we have:

`(a-1)(a-4)`

But `a=xy` so by substitution we have:

`(xy-1)(xy-4)`