Posted by **Tim** 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 -
**Anonymous**, Friday, April 18, 2008 at 6:12pm
No, I would not think you should have a fraction.

- Algebra -
**Reiny**, 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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