# Factorizing Fractions

1. Feb 27, 2008

### alpha01

1. The problem statement, all variables and given/known data

I dont understand why cross multiplying (a+1) with a(a+1) = (a+1)^2.

Similary on the RHS, I dont understand why cross multiplying (a-1) with a(a-1) = (a-1)^2.

2. Relevant equations

Factorizing to next step:

$$\frac{a + 1}{a(a - 1)}$$ _ $$\frac{a-1}{a(a + 1)}$$

gives:

$$\frac{(a + 1)^2}{a(a - 1)(a + 1)}$$ _ $$\frac{(a - 1)^2}{a(a - 1)(a + 1)}$$

Last edited: Feb 27, 2008
2. Feb 27, 2008

### rocomath

Property of exponents. If you have the same base/variable, and you're multiplying them, write it as one base/variable and just add their powers.

$$x\cdot x=x^1\cdot x^1=x^{1+1}=x^2$$

$$(x+1)\cdot(x+1)=(x+1)^1\cdot(x+1)^1=(x+1)^{1+1}=(x+1)^2$$

Also, your denominator contains a difference of squares: $$(a+1)(a-1)=a^2-1^2=a^2-1$$

Last edited: Feb 27, 2008
3. Feb 27, 2008

### Dick

You aren't 'cross multiplying', whatever that means. You are just putting things over a common denominator. Multiply the first term by (a+1)/(a+1)=1 and the second by (a-1)/(a-1)=1.

4. Feb 27, 2008

### alpha01

yes i understand that, however my question is what happened to the extra "a".

a(a+1)(a+1) = (a+1)^2?

which i know is not true

5. Feb 27, 2008

### rocomath

It doesn't disappear.

$$a(a+1)^2$$

Also, you don't need to multiply the other term by a. They both have a common term a. The first is missing a+1, and 2nd is missing a-1. That's all.

6. Feb 27, 2008

### alpha01

yes it does, please look at the solution above (its from my uni's course notes).

I have deleted "The attempt at a solution" which was just my attempt to remove confusion

7. Feb 27, 2008

### rocomath

Yeah it's right, it's done.