# Provide all details to decompose the fraction (x-4)/(x-2)(x-3) in partial fractions?

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### 1 Answer

We'll write the model of decomposing such a fraction into partial irreducible fractions;

(x-4)/(x-2)(x-3) = A/(x-2) + B/(x-3)

To compute the sum of the 2 fractions, they must have the same denominator. Since it is not the case, we'll create the same denominator to both elementary fractions.

(x-4) = A(x-3) + B(x-2)

We'll remove the brackets:

x-4 = Ax - 3A + Bx - 2B

We'll combine like terms:

x-4 = x(A+B) - 3A -2B

Comparing, we'll get:

According to this:

A+B=1 (1)

- 3A -2B = -4 (2)

We' ll multiply A+B=1 by 2, to eliminate the unknown B.

2A + 2B=2 (3)

Now, we'll add (3) to (2):

2A + 2B - 3A -2B = 2-4

We'll eliminate like terms:

-A = -2

A=2 => A+B=1 <=> 2+B=1 => B=1-2** => **B=-1

**The result of decomposition into partial fractions is: (x-4)/(x-2)(x-3) = 2/(x-2) - 1/(x-3).**