Complex Fractions & Negative Exponents

So I've set a goal for myself to relearn all the math I once knew, one textbook at a time. Realizing I could do at age 10 what I can't do now is humbling. Anyway.

I've got this problem and can't figure out how to get the correct answer:

c^{-2 }- 1

-------

c^{-1 }+ 1

I've tried to work it as follows:

1

-- - 1

c^{2 }^{---------
1
--- + 1 }c

Clear the fraction by multiplying by c^{2} across all numerators.

This leaves

1 - c^{2 }-------

c + c^{2}

However, my book says the answer is:

1 - c

------

c

I can't for the life of me get there.

Help?

[[EDIT:

Oh good lord. After trying for over an hour I worked backwards from the given solution and got how to simplify to the correct answer.

However, if this had been a proper exam I would've never thought to factor down to what is 'correct'. I only got there because I knew what was right.

Any tips on knowing when factoring is possible? Or quicker ways to recognize those patterns? Help would be appreciated.]]

Re: Complex Fractions & Negative Exponents

Hey pigeonpierce.

In terms of advice for factoring: try and reduce expressions down to the smallest factors possible.

The simplest factor possible is a multiplication of linear factors like (1+c)(1-c). If you can't get linear factors, then just go as far as you can towards linear.

The rest will be cancelling out factors leaving you with the minimal representation.

Re: Complex Fractions & Negative Exponents

Quote:

Originally Posted by **pigeonpierce;793798I've got this problem and can't figure out how to get the correct answer:**

c[SUP

-2 [/SUP]- 1

-------

c^{-1 }+ 1

We can factor this simply by using the *difference of two squares*.

$\displaystyle \frac{c^{-2}-1}{c^{-1}+1}=\frac{(c^{-1}-1)(c^{-1}+1)}{c^{-1}+1}=c^{-1}-1$

Re: Complex Fractions & Negative Exponents

Quote:

Originally Posted by

**pigeonpierce** So I've set a goal for myself to relearn all the math I once knew, one textbook at a time. Realizing I could do at age 10 what I can't do now is humbling. Anyway.

I've got this problem and can't figure out how to get the correct answer:

c^{-2 }- 1

-------

c^{-1 }+ 1

I've tried to work it as follows:

1

-- - 1

c^{2 }^{---------
1
--- + 1 }c

Clear the fraction by multiplying by c^{2} across all numerators.

This leaves

1 - c^{2 }-------

c + c^{2}

Plato has already indecated that you could use the "difference of squares" factoring directly from the original statement of the problem. You can also use it here:

$\displaystyle \frac{1- c^2}{c+ c^2}= frac{(1- c)(1+ c)}{c(1+ c)}= \frac{1- c}{c}$

since we can cancel the two "1+ c".

However, my book says the answer is:

1 - c

------

c

I can't for the life of me get there.

Help?

[[EDIT:

Oh good lord. After trying for over an hour I worked backwards from the given solution and got how to simplify to the correct answer.

However, if this had been a proper exam I would've never thought to factor down to what is 'correct'. I only got there because I knew what was right.

Any tips on knowing when factoring is possible? Or quicker ways to recognize those patterns? Help would be appreciated.]][/QUOTE]