# What steps should I take to solve this polynomial fraction problem?

• August 28th 2007, 02:01 PM
deathtolife04
What steps should I take to solve this polynomial fraction problem?
The problem is:

9c^7w^-4(-d^2)
______________
15c^3w^6(-d)^2

I can't remember what steps you're supposed to take for a problem like this? I'm especially confused by the negative power, and the d^2 and negative d^2.

Your input would be greatly appreciated! Thanks!
• August 28th 2007, 02:05 PM
Jhevon
Quote:

Originally Posted by deathtolife04
The problem is:

9c^7w^-4(-d^2)
______________
15c^3w^6(-d)^2

I can't remember what steps you're supposed to take for a problem like this? I'm especially confused by the negative power, and the d^2 and negative d^2.

Your input would be greatly appreciated! Thanks!

don't let the negatives bother you, just grind though. the negative in front of the d on the top is not being squared, so you can bring that out to the front. the one in front of the d at the bottom is being squared, so it will go away (that is, d^2 = (-d)^2), so you can pretty much forget about that one. otherwise, you need to know:

$\frac {x^a}{x^b} = x^{a - b}$

so you could write: $\frac {9c^7 w^{-4} (-d^2)}{15c^3w^6(-d)^2} = - \frac {9}{15} \cdot \frac {w^{-4}}{w^6} \cdot \frac {d^2}{d^2}$

where $\cdot$ means multiply.

Now continue
• August 28th 2007, 06:51 PM
deathtolife04
Quote:

Originally Posted by Jhevon
don't let the negatives bother you, just grind though. the negative in front of the d on the top is not being squared, so you can bring that out to the front. the one in front of the d at the bottom is being squared, so it will go away (that is, d^2 = (-d)^2), so you can pretty much forget about that one. otherwise, you need to know:

$\frac {x^a}{x^b} = x^{a - b}$

so you could write: $\frac {9c^7 w^{-4} (-d^2)}{15c^3w^6(-d)^2} = - \frac {9}{15} \cdot \frac {w^{-4}}{w^6} \cdot \frac {d^2}{d^2}$

where $\cdot$ means multiply.

Now continue

OK, so that would make my final answer: $\frac {-3} {5w^10}$

(to the power of ten on that last part, i couldn't figure out how to make the 1 AND the 0 small like that...)

right?

• August 28th 2007, 07:41 PM
Jhevon
Quote:

Originally Posted by deathtolife04
OK, so that would make my final answer: $\frac {-3} {5w^10}$

(to the power of ten on that last part, i couldn't figure out how to make the 1 AND the 0 small like that...)

right?

yes, that is correct

LaTex takes the first element after "^" and makes it a power. so typing ^10 causes LaTex to think 1 is the power and 0 is beside it. to get 10 as the power, make 10 a single element, that is, type "^{10}" ....these curly parenthesis are used for grouping in LaTex, when we want to identify two or more things as a single entity

EDIT: wait, we forgot something! what about the c?! It should be $\frac {9c^7 w^{-4} (-d^2)}{15c^3w^6(-d)^2} = - \frac {9}{15} \cdot \frac {c^7}{c^3} \cdot \frac {w^{-4}}{w^6} \cdot \frac {d^2}{d^2}$ ..........you should have caught that
• August 28th 2007, 08:04 PM
deathtolife04
Quote:

Originally Posted by Jhevon
yes, that is correct

LaTex takes the first element after "^" and makes it a power. so typing ^10 causes LaTex to think 1 is the power and 0 is beside it. to get 10 as the power, make 10 a single element, that is, type "^{10}" ....these curly parenthesis are used for grouping in LaTex, when we want to identify two or more things as a single entity

EDIT: wait, we forgot something! what about the c?! It should be $\frac {9c^7 w^{-4} (-d^2)}{15c^3w^6(-d)^2} = - \frac {9}{15} \cdot \frac {c^7}{c^3} \cdot \frac {w^{-4}}{w^6} \cdot \frac {d^2}{d^2}$ ..........you should have caught that

woah, good thing you caught that! i missed it cuz all my numbers were kinda squished together on my page. :eek:

So that would make the final answer: $\frac {-3c^4} {5w^{10}}$

right?
• August 28th 2007, 08:20 PM
Jhevon
Quote:

Originally Posted by deathtolife04
woah, good thing you caught that! i missed it cuz all my numbers were kinda squished together on my page. :eek:

So that would make the final answer: $\frac {-3c^4} {5w^{10}}$

right?

ok, now we're good! :)