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
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