# A couple of algebra problems...

• Nov 6th 2008, 08:29 AM
Grimes
A couple of algebra problems...
Hi, I'm new to the board.

I had a big ol' heaping pile of algebra homework to do recently, and I'm happy to say that I've actually figured most of it out, save for these four problems. It's really just two basic ideas that I don't understand, as the problems come in pairs.

For the first pair, I'm just asked to simplify:

a^2-2ab+b^2
______________

a-b

and...

a^2-2ab+b^2
______________

a+b

I've been trying to solve these using polynomial long division, but I don't think I'm doing the descending-power reordering right or something.

The second pair are just addition and subtraction problems, but I'm stumped. :p

a-1
____

a+1

+

a+1
____

a-1

and

4
____

a-5

-

1
____

5-a

If someone can help me understand how to solve these problems, I'd appreciate it.
• Nov 6th 2008, 09:23 AM
masters
Quote:

Originally Posted by Grimes
Hi, I'm new to the board.

I had a big ol' heaping pile of algebra homework to do recently, and I'm happy to say that I've actually figured most of it out, save for these four problems. It's really just two basic ideas that I don't understand, as the problems come in pairs.

For the first pair, I'm just asked to simplify:

$\displaystyle \frac{a^2-2ab+b^2}{a-b}$

First, factor the numerator.

$\displaystyle \frac{(a-b)(a-b)}{a-b}$

Then, divide out any common factors. In this case, $\displaystyle a-b$ is common in both the numerator and denominator. So, we end up with:

$\displaystyle \boxed{a-b}$

Quote:

Originally Posted by Grimes
$\displaystyle \frac{a^2-2ab+b^2}{a+b}$

Do this one the same way.

$\displaystyle \frac{(a-b)(a-b)}{a+b}$

There are no common factors here, so nothing cancels out. This cannot be simplified further.

Quote:

Originally Posted by Grimes

I've been trying to solve these using polynomial long division, but I don't think I'm doing the descending-power reordering right or something.

The second pair are just addition and subtraction problems, but I'm stumped. :p

$\displaystyle \frac{a-1}{a+1}+\frac{a+1}{a-1}$

We need to find the common denominator. Since they're both different the common denominator will be $\displaystyle (a+1 )(a-1)$

Now rewrite each fraction using the new denominator. We do this by multiplying this common denominator by each term; then put that result over the common denominator. Like this...

$\displaystyle \frac{(a+1 )(a-1)\left(\frac{a-1}{a+1}\right)+(a+1 )(a-1)\left(\frac{a+1}{a-1}\right)}{(a+1)(a-1)}=\frac{(a-1)(a-1)+(a+1)(a+1)}{(a+1)(a-1)}=$

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

Quote:

Originally Posted by Grimes

$\displaystyle \frac{4}{a-5}-\frac{1}{5-a}$

Let's rearrange that second fraction to get both denominators to look the same.

$\displaystyle \frac{4}{a-5}-\frac{1}{-(a-5)}=\frac{4}{a-5}+\frac{1}{a-5}=\frac{5}{a-5}$
• Nov 22nd 2008, 10:08 AM
Grimes
Thanks a lot for the help! :)