1. ## Binomial Factoring

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

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

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

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

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

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

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

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

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

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

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

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