Results 1 to 2 of 2

Math Help - Please check my answers- Partial Fractions

  1. #1
    Junior Member
    Joined
    Oct 2007
    Posts
    44

    Exclamation Please check my answers- Partial Fractions

    Write the partial fraction decomposition of the rational expression.

    3. (9x^2-x-14)/(x^3-x)

    I got: (14)/(x) +(-2)/(x+1) + (-3)/(x-1)


    Write the partial fraction decomposition of the rational expression.

    11. (6x+5)/((x-9)^2)

    I got: (6)/(x-9) + (59)/(x-9)^2


    13. (5x^2-5x+7)/((x-1)^3)
    I got: (5)/(x-1) - (5)/(x-1)^2 + (7)/(x-1)^2

    15. (x+4)/(x^3-2x^2+x)
    I got: (4)/(x) + (5)/(x-1) + (-4)/(x-1)^2

    Write the partial fraction decomposition of the rational expression.

    17. (12x+3)/ (x^3-1)
    I got: (-5)/(x-1) + (5x+2)/(x^2+x+1)

    (12x^2 -12x+6)/(x-3)(x^2+4)
    I got: (6)/(x-3) + (6x+6)/(x^2+4)

    Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants.

    26. (7x-3)/((x^2+x-7)^2)

    I got: (ax+b)/(x^2+x-7) + (cx+d)/((x^2+x+7)^2)

    29. (2x+1)/(x-6)((x^2+x-4)^2)

    I got: (a)/(x-6) + (bx+c)/(x^2+x-4)

    Write the partial fraction decomposition of the rational expression.


    31. (x^2+4x-1)/((x^2+3)^2)
    I got: (1)/(x^2-3) + (4x-4)/(x^2+3)^2

    33. (4x^3+ 4x^2)/((x^2+5)^2))
    I got: (4x+4)/(x^2+5) + (-20x-20)/((x^2+5)^2))
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,711
    Thanks
    630
    Hello, soly_sol!

    Write the partial fraction decomposition of the rational expression.

    3)\;\;\frac{9x^2-x-14}{x^3-x}

    I got: . \frac{14}{x} - \frac{2}{x+1} -\frac{3}{x-1} . . . . Right!

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    11)\;\;\frac{6x+5}{(x-9)^2}

    I got: . \frac{6}{x-9} + \frac{59}{(x-9)^2} . . . . Yes!

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    13)\;\;\frac{5x^2-5x+7}{(x-1)^3}

    I got: . \frac{5}{x-1} - \frac{5}{(x-1)^2} + \frac{7}{(x-1)^2} . . . . no

    It should be: . \frac{5}{x-1} + \frac{5}{(x-1)^2} + \frac{7}{(x-1)^3}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    15)\;\;\frac{x+4}{x^3-2x^2+x}

    I got: . \frac{4}{x} + \frac{5}{x-1} - \frac{4}{(x-1)^2} . . . . no

    It should be: . {\color{blue}\frac{4}{x} - \frac{4}{x-1} + \frac{5}{(x-1)^2}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    17)\;\;\frac{12x+3}{x^3-1}

    I got: . \frac{-5}{x-1} + \frac{5x+2}{x^2+x+1} . . . . no

    It should be: . {\color{blue}\frac{5}{x-1} + \frac{-5x+2}{x^2+x+1}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    You can check your answers yourself.
    Just combine the fractions in your answer.
    . . You should get the original fraction.

    In #13, they gave us: . \frac{5x^2 - 5x + 7}{(x-1)^3}

    You came up with: . \frac{5}{x-1} - \frac{5}{(x-1)^2} + \frac{7}{(x-1)^3}


    Let's check it . . . get a common denominator and combine.

    \frac{5}{x-1}\cdot{\color{green}\frac{(x-1)^2}{(x-1)^2}} \:- \:\frac{5}{(x-1)^2}\cdot{\color{green}\frac{x-1}{x-1}} \:+ \:\frac{7}{(x-1)^3}

    . . . = \;\frac{5(x-1)^2}{(x-1)^3} \:- \:\frac{5(x-1)}{(x-1)^3} \:+\: \frac{7}{(x-1)^3} \;\;=\;\;\frac{5(x-1)^2 - 5(x-1) + 7}{(x-1)^3}

    . . . = \;\frac{5x^2 - 10x + 5 - 5x + 5 + 7}{(x-1)^3} \;\;=\;\;\frac{5x^2 - {\color{red}15}x + {\color{red}17}}{(x-1)^3} ??

    This is not the original fraction . . . The solution is wrong.


    See how it works?

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Can someone check my answers?
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: April 15th 2010, 06:33 PM
  2. I want check my answers
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 2nd 2010, 07:13 PM
  3. Could someone check my answers pls.
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 6th 2009, 11:06 PM
  4. Check my answers please
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 15th 2009, 04:59 PM
  5. Partial Fractions- Please check my answers
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 24th 2007, 09:15 PM

Search Tags


/mathhelpforum @mathhelpforum