1. ## dividing polynomials

4x^-1y
10xy^-2

Now I know its 2/5 but I just dont know what to do with the negative exponents.

Another quick one.... (3/4)^-2
Is that just 9/8?

And I don't believe this is factorable... 36x^2 +49
Am I right?

2. Originally Posted by toddsayshi
4x^-1y
10xy^-2

Now I know its 2/5 but I just dont know what to do with the negative exponents.
Do you know negative exponents? $\displaystyle x^{-y}=\frac{1}{x^y}$ and $\displaystyle \frac{1}{x^{-y}}=x^y$

So you have: $\displaystyle \frac{4yx{-1}}{10xy^{-2}}$

I would split it up to be: $\displaystyle \frac{4y}{10x}\times x^{-1}\times \frac{1}{y^{-2}}$

Now get rid of the negative exponents: $\displaystyle \frac{4y}{10x}\times \frac{1}{x}\times y^2$

Multiply to get: $\displaystyle \frac{4y \times y^2}{10x\times x}$

Therefore: $\displaystyle \frac{4y^3}{10x^2}$

Which can be changed to: $\displaystyle \frac{2y^3}{5x^2}$

but that's as far as you can go.

Another quick one.... (3/4)^-2
Is that just 9/8?
$\displaystyle \left(\frac{3}{4}\right)^{-2}=\frac{3^{-2}}{4^{-2}}=\frac{4^2}{3^2}=\frac{16}{9}$

And I don't believe this is factorable... 36x^2 +49
Am I right?
I don't think it is, but 36x^2-49 is.

3. Here is a link that clears the negative exponents, it basically means something like this:

N^-x = 1/N^x

Math Forum - Ask Dr. Math Archives: Negative Exponents

On top of your problem there, is it 4x^-1 and then multiply by "y" or is it just 4x to the negative 1y power?

And the factoring, you can't factor that, but if in between it was a minus sign then you can: 36x^2-49 = (6x+7)(6x-7)

Hope this helps
Derg

4. Originally Posted by Dergyll
Here is a link that clears the negative exponents, it basically means something like this:

N^-x = 1/N^x

Math Forum - Ask Dr. Math Archives: Negative Exponents

On top of your problem there, is it 4x^-1 and then multiply by "y" or is it just 4x to the negative 1y power?

And the factoring, you can't factor that, but if in between it was a minus sign then you can: 36x^2-49 = (6x+7)(6x-7)

Hope this helps
Derg

It is 4x^-1 and then multiply by y.

5. Thats strange...I got the same answer with 2y^3/5x^2...are you sure it's 2/5 or are there other details?

Derg

6. Here it is....

7. Originally Posted by toddsayshi

Here it is....
Definitely $\displaystyle \frac{2y^3}{5x^2}$ not 2/5.

-Dan