Algebra Problem, help please?

  • I have lots of others like this, but I just want to figure out how to do it, so I can do the others myself.. Here's one example:



    Find the Partial Fraction decomposition of the Rational Function:

    (-2x^2 + 5x -1)/(x^4 - 2x^3 + 2x - 1)


    More details and guides would be appreciated ^^


  • First we need to factorise the denominator as far as it will go. We find that

    x^4 - 2x^3 + 2x - 1

    = (x + 1)(x^3 - 3x^2 + 3x - 1)

    = (x + 1)(x - 1)(x^2 - 2x + 1)

    = (x + 1)(x - 1)(x - 1)^2

    = (x + 1)(x - 1)^3, so

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

    (I did this by guessing likely roots (1 and -1) and using synthetic division.) So we need one term for each factor in the denominator, hence we let

    (-2x^2 + 5x -1)/(x + 1)(x - 1)^3 = A/(x + 1) + B/(x - 1) + C/(x - 1)^2 + D/(x - 1)^3.

    Now we multiply through by (x + 1)(x - 1)^3:

    -2x^2 + 5x -1 = A(x - 1)^3 + B(x + 1)(x - 1)^2 + C(x + 1)(x - 1) + D(x + 1).

    Now multiply out the RHS:

    -2x^2 + 5x -1 = A(x - 1)^3 + B(x + 1)(x - 1)^2 + C(x + 1)(x - 1) + D(x + 1)

    = Ax^3 - 3Ax^2 + 3Ax - A + Bx^3 - Bx^2 - 3Bx - B + Cx^2 - C + Dx + D.

    = (A + B)x^3 + (-3A - B + C)x^2 + (3A - 3B + D)x + (-A - B - C + D).

    Now we compare coefficients to get 4 equations in 4 unknowns.

    x^3 term: A + B = 0;
    x^2 term: -3A - B + C = -2;
    x term: 3A - 3B + D) = 5;
    constant term: -A - B - C + D = -1.

    Now we solve these any way we like. With more than 3 unknowns, I like to use matrices: let the matrix

    M = [[1, 1, 0, 0], [-3, -1, 1, 0], [3, -3, 0, 1], [-1, -1, -1, 1]]

    (written out row by row from the top) - compare with the LHS of the 4 equations - and vectors

    s = [[A], [B], [C], [D]] and

    v = [[0], [-2], [5], [-1]] - compare with the RHS.

    Then we have that Ms = v. So s = M^(-1)v. I use excel to calculate the inverse of M since it's hard :) and we get

    M^(-1) = (1/2)*[[1, -1, 1, -1], [1, 1, -1, 1], [-2, 6, -4, 4], [0, 6, -4, 6]] and

    M^(-1)v = [[4], [-4], [-18], [-19]] = s.

    So A = 4, B = -4, C = -18 and D = -19; so we have the result

    (-2x^2 + 5x -1)/(x^4 - 2x^3 + 2x - 1) = 4/(x + 1) - 4/(x - 1) - 18/(x - 1)^2 - 19/(x - 1)^3.

    For others like it, the process is the same; first factorise, and if you end up with (x + a)^m*(x + b)^n on the bottom, we want the original fraction to equal A/(x + a) + B/(x + a)^2 + C/(x + a)^3 + ... + J/(x + a)^m + K/(x + b) + L/(x + b)^2 + ... + Z(x + b)^n. Then multiply across by the deonminator, multiply out, compare co-efficients and solve the simultaneous equations. Hope that helped :)







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