How do you write the partial fraction decomposition of the rational expression (x+4)/((x+1)(x-2)^2 )x+4(x+1)(x−2)2?

1 Answer
Jim G.
Feb 12, 2016

1/3(x+1) - 1/3(x-2) + 2/(x-2)^2 13(x+1)−13(x−2)+2(x−2)2

Explanation:

firstly note that x-2)^2x−2)2
has factors (x-2) and (x-2)^2 (x−2)2

the factors of the denominator are linear hence the numerators are constants.

rArr (x+4)/((x+1)(x-2)^2 ) = A/(x+1) + B/(x-2) + C/(x-2)^2 ⇒x+4(x+1)(x−2)2=Ax+1+Bx−2+C(x−2)2

now multiply through by (x+1)(x-2)^2 (x+1)(x−2)2

hence : x+4 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)....................(1)

the aim now is to find values for A , B and C . Note that if x = 2 then the terms with A and B will be zero , and if x = -1 the terms with B and C will be zero. This is the starting point in finding values.

let x = 2 in (1) : 6 = 3C rArr C = 2

let x = -1 in (1) : -3 = 9A rArr A = 1/3
any value for x may be chosen to find B.

let x = 0 in (1) : 4 = 4A - 2B + C rArr 2B = 4A + C - 4

hence 2B = 4/3 + 2 - 4 = -2/3 rArr B = -1/3

rArr(x+4)/((x+1)(x-2)^2) = 1/3(x+1) -1/3(x-2) + 2/(x-2)^2

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