How do you write (3x)/((x + 2)(x - 1))3x(x+2)(x1) as a partial fraction decomposition?

1 Answer
Aug 4, 2016

= 2/(x+2) + 1/(x-1)=2x+2+1x1

Explanation:

(3x)/((x+2)(x-1)) = A/(x+2) + B/(x-1)3x(x+2)(x1)=Ax+2+Bx1

Getting every term over common denominator yields:

3x = A(x-1) + B(x+2)3x=A(x1)+B(x+2)

When I do these I eliminate one bracket by setting x to a particular value, but there are many other methods.

color(red)(x = 1: ) x=1:

3 = 3B implies B = 13=3BB=1

color(red)(x=-2: )x=2:

-6 = -3A implies A = 26=3AA=2

So (3x)/((x+2)(x-1)) = 2/(x+2) + 1/(x-1)3x(x+2)(x1)=2x+2+1x1