1. Binomial Factoring

$(x - y)^2 - 4(x - y) + 3$

I need some help with these kinds of binomials. I can never seem to get it right. I tried:

$(x - y)^2 - 4(x - y) + 3$

$[(x - y)^2 - 4(x - y)] + 3
=(x - y) [(x - y) - 4] - 3$

After that I don't know what to do, or I could have done it wrong already. Could anyone help please?

2. Hello,

Originally Posted by lax600
$(x - y)^2 - 4(x - y) + 3$

I need some help with these kinds of binomials. I can never seem to get it right. I tried:

$(x - y)^2 - 4(x - y) + 3$

$[(x - y)^2 - 4(x - y)] + 3
=(x - y) [(x - y) - 4] - 3$

After that I don't know what to do, or I could have done it wrong already. Could anyone help please?
Let $z=(x-y)$

So you want to factorise $z^2-4z+3=(z-3)(z-1)$

3. Can you factor $w^2 - 4w + 3$?
$w = \left( {x - y} \right)\quad \Rightarrow \quad \left( {x - y - 1} \right)\left( {x - y - 3} \right)
$

4. huh... wait wheres the z coming from?

5. Originally Posted by lax600
huh... wait wheres the z coming from?
Redefining the variable

It's easier to see the factorisation of zē-4z+3 than the original one.

After factorising, substitute back z by x-y

6. alright thx