What is partial fraction decomposition of 2x/(x^2-9)?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the partial fractions of 2x/(x^2-9)

2x/(x^2-9)

=> 2x/(x+3)(x - 3)

=> A/(x + 3) + B/(x - 3)

[A(x - 3) + B(x + 3)]/(x + 3)(x - 3) = 2x/(x+3)(x - 3)

=> Ax - 3A + Bx + 3B = 2x

A + B = 2 and 3B - 3A = 0

=> A + B = 2 and A = B

=> 2A = 2

=> A = 1 and as B = A = 1

The partial fractions are

1/(x + 3) + 1/(x - 3)

The partial fractions of 2x/(x^2-9) are 1/(x + 3) + 1/(x - 3)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write the fraction 2x/(x^2 - 9) as an algebraic sum of partial fractions: [A/(x-3)] + [B/(x+3)]

Since the LCD of the fractions from the right side is (x-3)(x+3) = x^2 - 9, we'll multiply by x^2 - 9 both fractions:

2x/(x^2 -9)= [A(x+3) + B(x-3)]/ (x^2 -9)

Having the common denominator (x^2 -9), we'll simplify it.

2x = Ax+3A+Bx-3B

We'll factorize by x to the right side:

2x = x*(A+B) + (3A-3B)

Comparing, we'll have:

A + B=2 (1)

3A-3B=0

We'll divide by 3:

A - B = 0 (2)

We'll add the second relation to the first one:

A + B + A - B=2+0

2A=2

A = 1

But, from (2) => A=B = 1

The partial fraction decomposition is: 2x/(x^2-9) = 1/(x-3) + 1/(x+3).