Finding partial fractions of (3x-2)/(x-3)(x+1):

Let us assume that (3x-2)/(x-3)(x+1) = A/(x-3) + B/(x+1)... (1)

We will detrmine the constants A and B :

WE multiply both sides by (x-3)(x+1) :

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

Put x = -1 in eq (2):

3(-1)-2 = A*0 +B(-1-3)

-5 = B(-4).

Divide by -4:

B = 5/4.

Put x = 3 in (2):

3(3)-2 = A(3+1).

7 = 4A.

A = 7/4.

Therefore substituting the value A = 7/4 and B = 5/4 in eq(1), we get:

(3x-2)/(x-3)(x+1) = 7/4(x-3)+5/4(x+1)

We'll decompose the given rational function in the elementary quotients:

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

We notice that the numerator of the original ratio is a linear function and the denominator is a quadratic function.

The irreducible ratios from the right side have as numerators constant functions and as denominators, linear functions.

We'll calculate LCD of the 2 ratios from the right side.

The LCD is the same with the denominator from the left side.

LCD = (x-3)(x+1)

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)

If the expressions from both sides are equal, then the correspondent coefficients are equals:

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

We'll divide by 4:

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

**(3x-2)/(x-3)(x+1) = 74/(x-3) + 5/4(x+1)**