What is the partial fraction decomposition of (5x+1)/(x^2+x-2) ?
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to find the partial fractions of (5x+1)/(x^2+x-2)
First let's factorize the denominator
x^2 + x - 2
=> x^2 + 2x - x - 2
=> x(x + 2) - 1(x + 2)
=> (x - 1)(x + 2)
(5x+1)/(x^2+x-2) = (5x+1)/(x - 1)(x + 2)
The partial fractions are A /(x - 1) + B/(x + 2)
A /(x - 1) + B/(x + 2) = (5x+1)/(x - 1)(x + 2)
=> A(x + 2) + B(x - 1) = 5x + 1
=> Ax + 2A + Bx - B = 5x + 1
=> 2A - B = 1 and A + B = 5
Add the two equations
=> 3A = 6
=> A = 2
B = 5 - A = 3
The partial fractions of (5x+1)/(x^2+x-2) are 2/(x - 1) + 3/(x + 2)
First thing, we'll write the denominator as a product of linear factors. For this reason, we'll determine it's roots.
We notice that the sum of roots is -1 and the product is -2, then the roots are x1 = 1 and x2 =- 2.
We can re-write the denominator as:
x^2 + x - 2 = (x-1)(x+2)
Since the factors from denominator are of the form (x - a), then we'll write the partial fractions as:
(5x + 1)/(x-1)(x+2) = A/(x-1) + B/(x+2)
(5x + 1) = A(x+2) + B(x-1)
We'll remove the brackets:
5x + 1 = Ax + 2A + Bx - B
5x + 1 = x(A+B) + 2A - B
Comparing, we'll get the system:
A + B = 5
2A - B = 1
Adding the equations above, we'll get:
A+B+2A-B = 5+1
3A = 6
A = 2
2 + B = 5 => B = 3
The complete partial fraction decomposition is (5x + 1)/(x-1)(x+2) = 2/(x-1) + 3/(x+2)