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.
- 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
-1 = A(48) + 0 ----> A = -1/48
3 = 16A + 4C
3 = 16(-1/48) + 4C ----> C = 5/6
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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