# Math Help - factoring help

1. ## factoring help

I'm struggling with factoring. Some are really easy, and some I don't know where to start.

m3 -mn2 +m2n - n3 + m2 - n2 is one that I need to factor.

The answer is (m+n)(m-n)(m+n+1) and if I multiply it all out, it gets me back to the original.

I get that (m+n)(m-n) is (m2 - n2). Do the substitution and you get m3 -mn2 +m2n - n3 + (m+n)(m-n).

You can factor m3 -mn2 +m2n - n3 into (m2-n2)(m+n), and further into (m+n)(m-n)(m+n).

Substitute again and you get (m+n)(m-n)(m+n) + (m+n)(m-n). Then I don't know where to go.

Can someone point me in the right direction?

Thanks

2. ## Re: factoring help

Originally Posted by baldysm
Substitute again and you get (m+n)(m-n)(m+n) + (m+n)(m-n). Then I don't know where to go.
You're so close! Just factorise the $(m+n)(m-n)$

$(m+n)(m-n)(m+n) + (m+n)(m-n)=(m+n)(m-n)\bigg((m+n)+1\bigg)$

3. ## Re: factoring help

Hmmm, lost me.

Are you factoring the (m+n)(m-n) from (m+n)(m-n)(m+n)?

Guess I don't see how to factor (m+n)(m-n) into (m+n+1)

4. ## Re: factoring help

Originally Posted by baldysm
Hmmm, lost me.
Are you factoring the (m+n)(m-n) from (m+n)(m-n)(m+n)?
No, I'm factoring $(m+n)(m-n)$ from the whole expression $(m+n)(m-n)(m+n) + (m+n)(m-n)$ .

I suppose it is easier to see if you use variables: Let $y=(m+n)(m-n)$. Then your expression becomes $y(m+n)+1y$. Factorising $y$ gives $y((m+n)+1)=(m+n)(m-n)(m+n+1)$

5. ## Re: factoring help

Ahhh, got it! Makes sense now.

Thanks!

6. ## Re: factoring help

Hello, baldysm!

I will change the variables.

$\text{Factor: }\:a^3 - ab^2 + a^2b - b^3 + a^2-b^2$

We have: . $(a^3-ab^2) + (a^2b - b^3) + (a^2-b^2)$

Factor: . $a(a^2-b^2) + b(a^2-b^2) + (a^2-b^2)$

Factor out $(a^2-b^2)\!:\;\;(a^2-b^2)(a + b + 1)$

Therefore: . $(a-b)(a+b)(a+b+1)$