# Thread: [Factorization] Factorization of Square factors

1. ## [Factorization] Factorization of Square factors

How do I factorize (further) this? I do not understand the steps, I know the answer although.

The question goes:

1)
(x-b)^2-x+b

=(x-b)^2-(x-b)
=(x-b)(x-b-1)

I do not understand the steps from 2nd to 3rd.

2)
x^2-y^2-(x-y)^2

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

I do not get the step from 1 to 2.

Explanation would be appreciated.

2. Originally Posted by Cthul
How do I factorize (further) this? I do not understand the steps, I know the answer although.

The question goes:

1)
(x-b)^2-x+b

=(x-b)^2-(x-b)
=(x-b)(x-b-1)

I do not understand the steps from 2nd to 3rd.

Explanation would be appreciated.
Suppose the equation is as such: $a^2-a$
To simplify it take $a$ common i.e $a(a-1)$

In the same way take $(x-b)$ common from $(x-b)^2-(x-b)$ i.e
$(x-b)[(x-b)-1]$.

You could also do it by substituting $a$ with $(x-b)$ in equation $a(a-1)=(x-b)[(x-b)-1]$

Hope this helps!!!

3. Originally Posted by Cthul
How do I factorize (further) this? I do not understand the steps, I know the answer although.

2)
x^2-y^2-(x-y)^2

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

I do not get the step from 1 to 2.

Explanation would be appreciated.

Do you know this identity $(a+b)(a-b)=a^2-b^2$

To solve this equation : $\color{red}x^2-y^2\color{black}-(x-y)^2$

$\color{red}(x+y)(x-y)\color{black}-(x-y)^2$

Hope this helps!!