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Question
Resolve the following rational expressions into partial fractions
`(x^2 + x + 1)/(x^2 - 5x + 6)`
Solution
`(x^2 + x + 1)/(x^2 - 5x + 6)`
Here the degree of the numerator is equal to the degree of the denominator.
Let us divide the numerator by the
∴ `(x^2 + x + 1)/(x^2 - 5x + 6) = 1 + (6x - 5)/(x^2 - 5x + 6)` ......(1)
Consider `(6x - 5)/(x^2 - 5x + 6)`
`(6x - 5)/(x^2 - 5x + 6) = (6x - 5)/(x^2 - 3x - 2x + 6)`
`(6x - 5)/(x^2 - 5x + 6) = (6x - 5)/(x(x - 3) - 2(x - 3))`
`(6x - 5)/(x^2 - 5x + 6) = (6x - 5)/((x - 2)(x - 3))`
`(6x - 5)/(x^2 - 5x + 6) = "A"/(x - 2) + "B"/(x - 3)` ......(2)
`(6x - 5)/(x^2 - 5x + 6) = ("A"(x - 3) + "B"(x - 2))/((x - 2)(x - 3))`
6x – 5 = A(x – 3) + B(x – 2) ......(3)
Put x = 3 in equation (3)
6(3) – 5 = A(3 – 3) + B(3 – 2)
18 – 5 = 0 + B
⇒ B = 13
Put x = 2 in equation (3)
6(2) – 5 = A(2 – 3) + B(2 – 2)
12 – 5 = – A + 0
7 = – A
⇒ A = – 7
Substituting the values of A and B in equation (2)
We have `(6x - 5)/(x^2 - 5x + 6) = (-7)/(x - 2) + 13/(x - 3)`
∴ The required partial fractions is
`(x^2 + x + 1)/(x^2 - 5x + 6) = 1 - 7/(x - 2) + 13/(x - 3)`
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