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Question
Resolve the following rational expressions into partial fractions
`(x + 12)/((x + 1)^2 (x - 2))`
Solution
`(x + 12)/((x + 1)^2 (x - 2)) = "A"/((x + 1)) + "B"/(x + 1)^2 + "C"/((x - 2))`
`(x + 12)/((x + 1)^2 (x - 2)) = ("A"(x + 1)(x + 2) + "B"(x - 2) + "C"(x + 1)^2)/((x + 1)^2 (x - 2)`
x + 12 = A(x + 1)(x – 2) + B(x – 2) + C(x + 1)2 ......(1)
Put x = 2 in equation (1)
2 + 12 = A(2 + 1)(2 – 2) + B(2 – 2) + C(2 + 1)2
14 = A(3)(0) + B × 0 + C(3)2
4 = 0 + 0 + 9C
⇒ C = `14/9`
Put x = – 1 in equation (1)
– 1 + 12 = A(– 1 + 1)(– 1 – 2) + B(– 1 – 2) + C(– 1 + 1)2
11 = A × 0 + B (– 3) + C × 0
11 = –3B
⇒ B = `- 11/3`
Put x = 0 in equation (1)
0 + 12 = A (0 + 1)(0 – 2) + B(0 – 2) + C(0 + 1)2
12 = – 2A – 2B + C
12 = `-2"A" - 2 xx - 11/3 + 14/9`
12 = `- 2"A" + 22/3 + 14/9`
2A = `22/3 + 14/9 - 12`
2A = `(66 + 14 - 108)/9`
= `(80 - 108)/9`
2A = `- 28/9`
⇒ A = `- 14/9`
∴ The required partial fraction is
`(x + 12)/((x + 1)^2 (x - 2)) = (- 14/9)/(x + 1) + (-11/3)/(x + 1)^2 + (14/9)/(x - 2)`
`(x + 12)/((x + 1)^2 (x - 2)) = - 14/(9(x + 1)) - 11/(3(x + 1)^2) + 14/(9(x - 2)`
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