Results 1 to 6 of 6

Math Help - What steps should I take to solve this polynomial fraction problem?

  1. #1
    Junior Member
    Joined
    Aug 2007
    Posts
    58

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by deathtolife04 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2007
    Posts
    58
    Quote Originally Posted by Jhevon View Post
    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?



    thanks for your help
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by deathtolife04 View Post
    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?



    thanks for your help
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2007
    Posts
    58
    Quote Originally Posted by Jhevon View Post
    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.

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

    right?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by deathtolife04 View Post
    woah, good thing you caught that! i missed it cuz all my numbers were kinda squished together on my page.

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

    right?
    ok, now we're good!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: November 13th 2011, 05:30 PM
  2. STEPS to Solve simple equation
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 13th 2010, 08:41 AM
  3. Replies: 2
    Last Post: March 29th 2010, 09:52 PM
  4. Steps to solving polynomial + example
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: October 27th 2008, 07:33 PM
  5. Need steps how to solve this little problem!
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: July 21st 2008, 01:07 PM

Search Tags


/mathhelpforum @mathhelpforum