Results 1 to 4 of 4

Math Help - hELP partial fractions....

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    hELP partial fractions....

    III. Distinct Quadratic Factors
    Resolve into partial...
    4.) (3x^3 + 8x^2 + 6x - 2) / (x+2)(x+1)(x^2+2)
    And the rule
    = A / (x +2) + B ( x+1) + (Cx + D)/(x^2 + 2)

    the only part i cant seem to find is by solving A,B,C and D
    3 = A + B + C
    8 = A + 2B + 3C + D
    6 = 2A + 2B + 2C + 3D
    -2 = 2A + 4B + 2D
    can u check if thiss correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    III. Distinct Quadratic Factors
    Resolve into partial...
    4.) (3x^3 + 8x^2 + 6x - 2) / (x+2)(x+1)(x^2+2)
    And the rule
    = A / (x +2) + B ( x+1) + (Cx + D)/(x^2 + 2)

    the only part i cant seem to find is by solving A,B,C and D
    3 = A + B + C
    8 = A + 2B + 3C + D
    6 = 2A + 2B + 2C + 3D
    -2 = 2A + 4B + 2D
    can u check if thiss correct?
    These give the correct solution, so yes they are correct.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    cud u help me answer please that is the part where i got stuck
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    III. Distinct Quadratic Factors
    Resolve into partial...
    4.) (3x^3 + 8x^2 + 6x - 2) / (x+2)(x+1)(x^2+2)
    And the rule
    = A / (x +2) + B ( x+1) + (Cx + D)/(x^2 + 2)

    the only part i cant seem to find is by solving A,B,C and D
    3 = A + B + C
    8 = A + 2B + 3C + D
    6 = 2A + 2B + 2C + 3D
    -2 = 2A + 4B + 2D
    can u check if thiss correct?
    Look at equations 1 and 3, one is twice the other except for an extra
    3D, so D must be zero.

    So the equations now become (after discarding the duplicate)

    3 = A + B + C
    8 = A + 2B + 3C
    -2 = 2A + 4B

    multiply equation 2 by 2 and we have:

    16=2A+4B+6C

    but from wuation 3 we have

    -2=2A+4B,

    so C=18/6=3.

    so now we have:

    A+B=0
    2A+4B=-2

    so 2B=-2, B=-1, A=1.

    Now you will need to check that this is correct as my algebra comes with
    no guarantees

    RonL

    RonL
    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