Question

Write out the form of the partial fraction decomposition of the function (See Example). Do not determine the numerical values of the coefficients. (If the partial fraction decomposition does not exist, enter DNE.)
(a) (x)/(x^2 + x − 6)

Answer 1

I have a solution here for the same problem but a different given:

(2x-82)/(x^2+2x-48) dx

In which this solution will help you answer the problem by yourself:

(2x - 82) /(x² + 2x - 48) 

first, factor the denominator completely: 

(x² + 2x - 48) = 

(x² + 8x - 6x - 48) = 

x(x + 8) - 6(x + 8) = 

(x - 6)(x + 8) 

so the function becomes: 

(2x - 82) /[(x - 6)(x + 8)] 

let's decompose this into partial fractions: 

(2x - 82) /[(x - 6)(x + 8)] = A/(x - 6) + B/(x + 8) 

(letting [(x - 6)(x + 8)] be the common denominator at the right side too) 

(2x - 82) /[(x - 6)(x + 8)] = [A(x + 8) + B(x - 6)] /[(x - 6)(x + 8)] 

(equating numerators) 

2x - 82 = Ax + 8A + Bx - 6B 

2x - 82 = (A + B)x + (8A - 6B) 

yielding the system: 

A + B = 2 
8A - 6B = - 82 

A = 2 - B 
4A - 3B = - 41 

A = 2 - B 
4(2 - B) - 3B = - 41 

A = 2 - B 
8 - 4B - 3B = - 41 

A = 2 - B 
- 7B = - 41 - 8 

A = 2 - B 
7B = 49 

A = 2 - 7 = - 5 
B = 49/7 = 7 

hence: 

(2x - 82) /[(x - 6)(x + 8)] = A/(x - 6) + B/(x + 8) = - 5/(x - 6) + 7/(x + 8) 

thus the answer is: 


(2x - 82) /(x² + 2x - 48) = [- 5 /(x - 6)] + [7 /(x + 8)]