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

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember the technique called partial fraction decomposition such that:

(3x-2)/(x-3)(x+1) = a/(x-3) + b/(x+1)

Notice that the numerators are not known yet, hence two variables are assigned.

You may get rid of all denominator multiplying all terms by the common denominator (x-3)(x+1) such that:

3x-2 = a(x+1) + b(x-3)

Opening the brackets yields: 3x-2=ax+a+bx-3b

Group the x terms and the constant terms such that: 3x-2=(a+b)x + (a-3b)

Equating the coefficients of like powers yields:

a+b = 3 => a = 3-b

a - 3b = -2 => 3 - b - 3b = -2 => -4b = -5 => b = 5/4

a = 3 - 5/4 = 7/4

Hence, the original fractions were the following: (3x-2)/(x-3)(x+1) = 7/(4x-12) + 5/(4x+14)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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)

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)