# Thread: Factorsing by removing a common factor

1. ## Factorsing by removing a common factor

I can factorize problems like this:
3x^2 + 5x
3x^2: (1) (3) (x) (x)
5x: (1) (5) (x)
Using 1 and x to drop outside and use the leftovers:
x(3x + 5) *final*

But I'm being stumped by this kind of problem:
(x + 2)^2 - 5(x + 2) ?

2. Originally Posted by robbz
But I'm being stumped by this kind of problem:
(x + 2)^2 - 5(x + 2) ?
First expand both sets of brackets:

$(x + 2)^2 - 5(x + 2) = x^2+4x+4-5x-10$

Collect like terms:

$x^2-x-6$

Then factorise like you would a normal quadratic equation (I am assuming that you have come across these before, if not please say).

Hope this helps.

3. Ok, so when you left off at x^2 - x - 6.

How do I factorize that? It's longer than the binomials I've been doing.

4. Think about it like this:

$
(x + 2)^2 - 5(x + 2) ~=~ (x+2)\textcolor{red}{(x+2)} - (5)\textcolor{red}{(x+2)}
$

So, we see that $(x+2)$ is common in both terms, so we factor the whole thing out:

$\textcolor{red}{(x+2)} \left[ (x+2) - 5 \right]$

Then you can simplify the right half.

5. Robbz, drumist's response is the one to go with!

6. Thanks for the help everyone.
I understand it now.