Use partial fractions(3x-2)/(x-3)(x+1)
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)