Factor out a common term

Question

I have this fraction, and I'm trying to simplify it. It's obvious that we have (x+1) as a common term.

$\frac{(x+1)^2-(x+2)[2(x+1)]}{(x+1)^4}$

In the next step it should look like: $\frac{(x+1)[x+1-2x-4]}{(x+1)^4}$

The thing which confuses me is that in the non simplified fraction (x+1) occurs three times, and when factored out it occurs only twice. How exactly do I get to the simplified fraction posted above?

I've tried writing it like:

$\frac{(x+1)(x+1)-(x+2)[2(x+1)]}{(x+1)^4}$

But I still don't understand how do we lose one (x+1) in the next step

Answer 1

Your numerator simplifies to $$(x+1)(-x-3)$$ and for $x\ne -1$ we get $$\frac{-x-3}{(x+1)^3}$$ or $$-\frac{x+3}{(x+1)^3}$$

Answer 2

Note that by the commutative property of multiplication, we have $$\color{red}{(x+1)}(x+1)-(x+2)[2\color{red}{(x+1)}]=\color{red}{(x+1)}(x+1)+\color{red}{(x+1)}\cdot (-2(x+2))$$ By the distributive property, we have that $$\color{red}{(x+1)}(x+1)+\color{red}{(x+1)}\cdot (-2(x+2))=\color{red}{(x+1)}\cdot((x+1)-2(x+2))$$ which we can simplify to $$\color{red}{(x+1)}\cdot((x+1)-2(x+2))=\color{red}{(x+1)}\cdot((x+1)-2x-4)=\color{red}{(x+1)}\cdot(-x-3)$$ So we end up with $$\frac{\color{red}{(x+1)}[x+1-2x-4]}{(x+1)^4}=\frac{\color{red}{(x+1)}\cdot(-x-3)}{\color{red}{(x+1)}(x+1)^3}=-\frac{x+3}{(x+1)^3}$$