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)
No, I would not think you should have a fraction.
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.
Let's go through the steps of finding the unknowns A, B, and C.
Step 1: Multiply both sides of the equation by the denominator on the right side to clear the fractions:
(x+3)/(x^3+64) = A/(x+4) + (Bx+C)/(x^2-4x+16)
Multiply both sides by (x^2-4x+16) to get rid of the denominators on the right side:
(x+3)(x^2-4x+16) = A(x^2-4x+16) + (Bx+C)(x+4)
Step 2: Expand and simplify the equation:
x^3 - x^2 + 12x + 48 = Ax^2 - 4Ax + 16A + Bx^2 + 4Bx + Cx + 4C
Step 3: Collect like terms on both sides:
x^3 + (-1 + B)x^2 + (12 + 4B + C)x + (48 + 16A + 4C) = Ax^2 - 4Ax + 16A
Step 4: Equate the coefficients of like powers of x. This will give you a system of equations to solve for A, B, and C:
Coefficient of x^3: 1 = 0A
Coefficient of x^2: -1 + B = A
Coefficient of x: 12 + 4B + C = -4A
Coefficient of x^0 (constant term): 48 + 16A + 4C = 0
Step 5: Solve the system of equations to find the values of A, B, and C.
Using the first equation, A = 0.
Substituting A = 0 into the second equation, -1 + B = 0, B = 1.
Substituting A = 0 and B = 1 into the third equation, 12 + 4(1) + C = -4(0), C = -16.
Substituting A = 0 and C = -16 into the fourth equation, 48 + 16(0) + 4(-16) = 0.
Therefore, the unknowns are A = 0, B = 1, and C = -16.
To recap, using the system of equations method, you obtained A = 0, B = 1, and C = -16 as the values for the unknowns in the partial fraction decomposition.