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