Resolve 3x^3-x^2-13x-13/x^2-x-6 into partial fractions
We want to resolve the given rational function into partial fractions:
3x^3 - x^2 - 13x - 13 / (x^2 - x - 6)
First, factor the denominator:
x^2 - x - 6 = (x - 3)(x + 2)
The partial fractions decomposition will have the form:
(3x^3 - x^2 - 13x - 13) / ((x - 3)(x + 2)) = A / (x - 3) + B / (x + 2)
To find A and B, we clear the denominators:
3x^3 - x^2 - 13x - 13 = A(x + 2) + B(x - 3)
Now we can find the values of A and B by substitution:
1. Choose x = 3:
3(3)^3 - (3)^2 - 13(3) - 13 = A(3 + 2)
63 = 5A
A = 12.6
2. Choose x = -2:
3(-2)^3 - (-2)^2 - 13(-2) - 13 = B(-2 - 3)
-13 = -5B
B = 2.6
Now we have the partial fractions:
(3x^3 - x^2 - 13x - 13) / ((x - 3)(x + 2)) = 12.6 / (x - 3) + 2.6 / (x + 2)