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Posted by on Friday, April 18, 2008 at 5:48pm.

Below is my problem with my answers. I don't think this is that tough, but I cannot believe that my ABC answer's are a fraction. Did I do this correctly?

Find the unknowns on the right side of the partial fraction.

(x+3)/(x^3+64)=(A)/(x+4)+(Bx+C)/(x^2-4x+16)

x=0
A= (-x^2B-4xB-x+x-4C+3)/(x^2-4x+16)
B= (-x^2A+4xA-x(-16A+x-4C+3)/(x(x+4)
C= (-x^2B-4xB-x^2A+4x+x-16A+3)/(x+4)

  • Algebra - , Friday, April 18, 2008 at 6:12pm

    No, I would not think you should have a fraction.

  • Algebra - , Friday, April 18, 2008 at 6:19pm

    I multiplied both sides of your equation by (x+4)(x^2 - 4x + 16) which is really x^3 + 64

    x+3 = A(x^2 - 4x + 16) + (x+4)(Bx+C)

    this is an identity, so it is true for any value of x.
    Let's pick "easy" values of x

    let x=-4
    -1 = A(48) + 0 ----> A = -1/48

    let x=0

    3 = 16A + 4C
    3 = 16(-1/48) + 4C ----> C = 5/6

    let x=1
    4 = 13A + 5B + 5C
    4 = 13(-1/48) + 5B + 5(5/6)
    B = 1/48

    Check my arithmetic, my weak point.

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