Express (3x^2 - 3x - 2)/(x-1)(x-2) in partial fractions

Step 1:

When asked to exrpess something in partial fractions, we first compare the power of the numerator to the power of the denominator.

In our case we have that the power of quadratic equation in the numerator is equal to 2, while the power of the denominator is 

(x-1)(x-2) = x- 3x +2 

which is equal to 2 as well.

Step 2:

Now we devide the numerator by the denominator.

Using long division we get that

(3x2 - 3x - 2)/(x-1)(x-2) = 3 + (6x-8)/(x- 3x + 2)

Step 3:

The next step is express (6x-8)/(x- 3x + 2) as a partial fraction

(6x-8)/(x- 3x + 2) = (6x-8)/(x-1)(x-2) 

(6x-8)/(x-1)(x-2) = A/x-1 + B/x-2

Step 4:

Now we multiply both the LHS and the RHS by (x-1)(x-2) because this leads to a common denominator.

6x - 8 = A(x-2) + B(x-1)

Now we have to use two different values for x, such that in the first instance B=0, and in the second instance,  A=0

Hence, when x=1,

6(1) - 8 = A(1-2) + B(1-1)

6 - 8 = -A + 0

-2 = - A

A = 2

When x=2

6(2) - 8 = A(2-2) + B(2-1)

12 - 8 = 0 + B

B = 4

Step 5:

Now going back to our original equation,

(3x2 - 3x - 2)/(x-1)(x-2) = 3 + (6x-8)/(x- 3x + 2)

 3 + (6x-8)/(x- 3x + 2) = 3 +  A/x-1 + B/x-2

and using A=2 , B=4 we get

3 + (6x-8)/(x- 3x + 2) = 3 +  2/x-1 + 4/x-2

Hence, our desired result is

(3x2 - 3x - 2)/(x-1)(x-2) =  3 +  2/x-1 + 4/x-2

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