# What is the partial fraction decomposition of (5x+1)/(x^2+x-2) ?

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### 2 Answers

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)**