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Math Help - Partial Fractions

  1. #1
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    Exclamation Partial Fractions

    Partial Fraction Decomposition
    Will you please help me with the following problems:

    Write the partial fraction decomposition of the rational expression.

    x/(x^2-5x+6)

    (7x^2-x-12)/x(x+1)(x-1)

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

    Write the partial fraction decomposition of the rational expression.

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

    (5x^2-5x+7)/(x-1)^3

    (x+4)/(x^3-2x^2+x)


    Write the partial fraction decomposition of the rational expression.

    (12x+3)/(x^3-1)

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

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

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

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

    Write the partial fraction decomposition of the rational expression.

    (x^2+4x-1)/(x^2+3)^2

    (4x^3+4x^2)/(x^2+5)^2
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by soly_sol View Post
    Partial Fraction Decomposition
    Will you please help me with the following problems:

    Write the partial fraction decomposition of the rational expression.

    x/(x^2-5x+6)

    (7x^2-x-12)/x(x+1)(x-1)

    (9x^2-x-14)/(x^3-x)
    The trick is:
    If you have a factor of ax + b in the denominator you have a term \frac{A}{ax + b} in your decomposition.

    If you have a factor of ax^2 + bx + c in the denominator you have a term \frac{Ax + B}{ax^2 + bx + c} in your decomposition.

    etc.

    If you have factor of (ax + b)^2 in your denominator then you have terms \frac{A}{ax + b} + \frac{B}{(ax + b)^2} in your decomposition.


    If you have factor of (ax + b)^3 in your denominator then you have terms \frac{A}{ax + b} + \frac{B}{(ax + b)^2} + \frac{C}{(ax + b)^3} in your decomposition.

    etc.

    As an example, let's do the first one:
    \frac{x}{x^2-5x+6} = \frac{x}{(x -2)(x - 3)}

    We have linear factors of x - 2 and x - 3 in the denominator, so we use
    \frac{x}{(x -2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}

    How do we find A and B? My favorite method (there are many) is to add the fractions on the LHS:
    \frac{x}{(x -2)(x - 3)} = \frac{A(x - 3) + B(x - 2)}{(x - 2)(x - 3)} = \frac{(A + B)x + (-3A - 2B)}{(x - 2)(x - 3)}
    and match the coefficients of powers of x on both sides of the equation.

    So we know that
    A + B = 1
    and
    -3A - 2B = 0

    I get that A = -2 and B = 3, thus
    \frac{x}{x^2-5x+6} = -\frac{2}{x - 2} + \frac{3}{x - 3}

    It might take longer than some of the other tricks, but I find it keeps me more organized with what I am doing.

    Try the other 2 in this set. I got:
    \frac{7x^2-x-12}{x(x+1)(x-1)} = \frac{12}{x} - \frac{3}{x - 1} - \frac{2}{x + 1}
    and
    \frac{9x^2-x-14}{x^3-x} = \frac{14}{x} - \frac{3}{x - 1} - \frac{2}{x + 1}

    -Dan
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  3. #3
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    Quote Originally Posted by soly_sol View Post
    x/(x^2-5x+6)
    Most of the times I try to avoid partial fractions

    Here's a slightly variation:

    x^2-5x+6=(x-2)(x-3).

    We have to manipulate the numerator such that this be equal to x. Note that (tricky)

    3(x-2)-2(x-3)=3x-6-2x+6=x. So, with this we're are done, and

    \frac{x}<br />
{{x^2 - 5x + 6}} = \frac{{3(x - 2) - 2(x - 3)}}<br />
{{(x - 2)(x - 3)}} = \frac{3}<br />
{{x - 3}} - \frac{2}<br />
{{x - 2}},

    which confirms Dan's answer.
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  4. #4
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    (7x^2-x-12)/x(x+1)(x-1)

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

    Can you please explain to me how to these? I am having some trouble starting out. For the first one could I make the denomintator x(x-1)^2 since its a difference of squares?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by soly_sol View Post
    (7x^2-x-12)/x(x+1)(x-1)
    set \frac {7x^2 - x - 12}{x(x + 1)(x - 1)} = \frac Ax + \frac B{x + 1} + \frac C{x - 1}


    (9x^2-x-14)/(x^3-x)
    set \frac {9x^2 - x - 14}{x^3 - x} = \frac {9x^2 - x - 14}{x \left( x^2 - 1 \right)} = \frac {9x^2 - x - 14}{x(x + 1)(x - 1)} = \frac Ax + \frac B{x + 1} + \frac C{x - 1}
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  6. #6
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    Am I doing this right so far:

    for the first one: 7x^2-x-12=a(x+1)(x-1)+Bx(x-1)+Cx(x+1)

    then for the second one: 9x^2-x-14=a(x+1)(x-1)+Bx(x-1)+Cx(x+1)
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by soly_sol View Post
    Am I doing this right so far:

    for the first one: 7x^2-x-12=a(x+1)(x-1)+Bx(x-1)+Cx(x+1)

    then for the second one: 9x^2-x-14=a(x+1)(x-1)+Bx(x-1)+Cx(x+1)
    yes
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