The partial fractions can be found in a very easy way using the following method.
(3x-2)/(x-3)(x+1), the roots of the denominator are x = 3 and x = -1.
Representing as partial fractions we get the form A/(x - 3) + B(x + 1)
Let's find A, ignore the term (x - 3) and substitute the root x = 3 for all the other terms.
This gives us (3*3 - 2)/(3 + 1) = (9 - 2)/4 = 7/4
Similarly to find B, eliminate (x + 1) and substitute the root x = -1 in the other terms, we get (3*-1 - 2)/(-1 - 3) = 5/4
Therefore the partial fractions are (7/4)/(x - 3) + (5/4)/(x + 1)
so now for simplifying fractions like this we break this in 2 parts
(3x-2)/(x-3)(x+1)=a/(x-3) + b/(x+1)
now put x=3
4a=7 or a=7/4
also now put x=-1
so now put these all above these are your partial fractions.
We'll decompose the rational function in partial fractions:
(3x-2)/(x-3)(x+1) = A/(x-3) + B/(x+1)
We'll calculate LCD of the 2 fractions from the right side.
The LCD is the same with the denominator from the left side.
LCD = (x-3)(x+1)
We'll multiply both sides by (x-3)(x+1) and the expression will become:
(3x-2) = A(x+1) + B(x-3)
We'll remove the brackets:
3x - 2 = Ax + A + Bx - 3B
We'll combine like terms form the right side:
3x - 2 = x(A+B) + (A-3B)
Comparing we'll get:
3 = A+B
-2 = A - 3B
We'll use the symmetric property:
A+B = 3 (1)
A - 3B = -2 (2)
We'll multiply (1) by 3:
3A+3B = 9 (3)
We'll add (3) to (2):
3A+3B+A - 3B = 9-2
We'll eliminate like terms:
4A = 7
A = 7/4
We'll substitute A in (1):
A+B = 3
7/4 + B = 3
B = 3 - 7/4 => B = (12-7)/4 => B = 5/4
The fraction (3x-2)/(x-3)(x+1) = 7/4(x-3) + 5/4(x+1), where 7/4(x-3) and 5/4(x+1) are partial fractions.