Results 1 to 9 of 9

Math Help - Partial Fractions Help

  1. #1
    Member
    Joined
    Dec 2007
    From
    Georgia
    Posts
    85

    Partial Fractions Help

    Ok. I gotta do the fraction decomposition of

     \frac{6x^2+1}{x^2(x-1)^3}.

    ~~~~~~~~~~~

    My problem is when I get to the part where I figure out what goes into A-E. I'll show the work I have right now below.

    6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2
    6x^2+1=A(x^4+3x^3+3x^2-x)+B(x^3 + x^2 + 3x-1)+C(x^4-2x^3+x^2)+D(x^3-x^2)+Ex^2
    6x^2+1=(A+C)x^4+(3A+B-2C+D)x^3+(3A+3B+C-D+E)x^2+(-A-B)x+(-B)

    Given the highest power of the numerator is x^2. I'm really at a lost what to do from here. I'm not even sure if I'm doing it right. So some help would be greatly appreciated

    EDIT: I'm gonna go rework this, something tells me I'm really close to the answer.
    Last edited by mathgeek777; March 22nd 2008 at 12:28 PM. Reason: Informing
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    \frac{7}{(x-1)^{3}}-\frac{2}{(x-1)^{2}}+\frac{3}{x-1}-\frac{3}{x}-\frac{1}{x^{2}}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mathgeek777 View Post
    Ok. I gotta do the fraction decomposition of

     \frac{6x^2+1}{x^2(x-1)^3}.

    ~~~~~~~~~~~

    My problem is when I get to the part where I figure out what goes into A-E. I'll show the work I have right now below.

    6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2
    6x^2+1=A(x^4+3x^3+3x^2-x)+B(x^3 + x^2 + 3x-1)+C(x^4-2x^3+x^2)+D(x^3-x^2)+Ex^2
    6x^2+1=(A+C)x^4+(3A+B-2C+D)x^3+(3A+3B+C-D+E)x^2+(-A-B)x+(-B)

    Given the highest power of the numerator is x^2. I'm really at a lost what to do from here. I'm not even sure if I'm doing it right. So some help would be greatly appreciated

    EDIT: I'm gonna go rework this, something tells me I'm really close to the answer.
    6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2


    Since it's gotta work for all values of x, you can substitute at least one simple value of x to make life slightly easier:

    x = 0: 1 = -B therefore B = -1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2007
    From
    Georgia
    Posts
    85
    Quote Originally Posted by mr fantastic View Post
    6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2


    Since it's gotta work for all values of x, you can substitute at least one simple value of x to make life slightly easier:

    x = 0: 1 = -B therefore B = -1.
    wait, what?

    you lost me.....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by mathgeek777 View Post
    wait, what?

    you lost me.....

    What Mr. Fantastic is saying is that the Equation is true for ALL values of x.

    so you can pick some to make your computations easier. Like x=0

    <br />
6(0)^2+1=A[0(0-3)]^3+B(0-1)^3+C[0^2(0-1)^2]+D[0^2(x-1)]+E(0)^2<br />

    Simplifying

    1=-B
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    Since it's gotta work for all values of x, you can substitute at least one simple value of x to make life slightly easier:

    x = 0: 1 = -B therefore B = -1.
    6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2

    is an identity, true for all values of x.

    So it has to be true in particular for convenient simple values of x like x = 0. So substitute x = 0:

    The left hand side becomes 6(0)^2 + 1 = 1.
    The right hand side becomes B(-1)^3 = -B.

    This idea can sometimes be a good short cut for getting the values of A, B etc. rather than expanding, equating coefficients, writing down and then solving simultaneous equations etc.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2007
    From
    Georgia
    Posts
    85
    Quote Originally Posted by mr fantastic View Post
    6x^2+1=A[x(x-3)]^3+B(x-1)^3+C[x^2(x-1)^2]+D[x^2(x-1)]+Ex^2

    is an identity, true for all values of x.

    So it has to be true in particular for convenient simple values of x like x = 0. So substitute x = 0:

    The left hand side becomes 6(0)^2 + 1 = 1.
    The right hand side becomes B(-1)^3 = -B.

    This idea can sometimes be a good short cut for getting the values of A, B etc. rather than expanding, equating coefficients, writing down and then solving simultaneous equations etc.
    Can this be done successfully with the rest of the missing values on partial fractions?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Dec 2007
    From
    Georgia
    Posts
    85
    My computer froze, please disregard
    Last edited by mathgeek777; March 22nd 2008 at 05:47 PM. Reason: computer froze
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mathgeek777 View Post
    Can this be done successfully with the rest of the missing values on partial fractions?
    Yes, but you'll get a series of equations to be solved simultaneously rather than any further values dropping out easily.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: April 28th 2010, 09:53 AM
  2. Partial Fractions
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 4th 2010, 06:24 PM
  3. Partial Fractions
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: January 17th 2010, 02:20 AM
  4. Help on Partial Fractions
    Posted in the Calculus Forum
    Replies: 12
    Last Post: January 6th 2010, 03:00 AM
  5. partial fractions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 21st 2007, 04:49 AM

Search Tags


/mathhelpforum @mathhelpforum