partial fraction decomposition of (x+2)/(x^3-2x^2+x)?

Answer 1

denominator = x^3-2x^2+x = x(x^2-2x+1) = x(x-1)^2

so we get (x+2)/(x^3-2x^2+x) = A/x + B/(x-1) + C/(x-1)^2
we can proceed and find

Answer 2

first recognize that the denominator is x*(x-1)^2 then this factors to;

A/x + B/(x-1) + C/(x-1)^2 so

(x+2)/x(x-1)^2 = A/x + B/(x-1) + C/(x-1)^2
then multiply through by the denominator on the left side of the equation to get;

(x+2) = A(x-1)^2 + Bx(x-1) + Cx or
(x+2) = A(x^2-2x+1) + B(x^2-x) + Cx or
(x+2) = (Ax^2 -2Ax + A) + (Bx^2 – Bx) + Cx or
(x+2) = (A+B)x^2 + (-2A-B+C)x + A

so for this equality to work, the coefficients on the left hand side of the equation must equal the coefficients on the right hand side of the equation or

A=2 (the constant values) so this makes A+B = 2+B = 0 (the coefficient of x^2 on the left hand side of the equation is 0 since there is no x^2 on the left hand side of the equation) or B = -2
and now -2A-B+C = 1 (the coefficient of x on the left hand side of the equation) or -2(2) -(-2) + C = 1 or -4+2 + C = 1 or -2 + C = 1 or C = 1+2 = 3

so your answer is 2/x - 2/(x-1) + 3/(x-1)^2

please double check my algebra because I’m late for work now! Good luck!

Answer 3

x^3-2x^2+x
=x(x^2-2x+1)
=x(x-1)^2.
(x+2)/(x^3-2x^2+x)=(x+2)/x(x-1)^2
let
(x+2)/x(x-1)^2=A/x + B/(x-1)+C/(x-1)^2
multiply both sides by x(x-1)^2;
x+2=A(x-1)^2+Bx(x-1)+Cx;
Put x=0;
2=A, A=2;
Put x=1;
3=C, C=3;
compare the coefficient of x^2;
0=A+B, B=-2
so the desired P.F 's are 2/x-2/(x-1)+3/(x-1)^2 ANS

Answer 4

x+2 / x(x^2 -2x +1)

x+2 / x(x-1)(x-1)

[a / x] + [b / (x-1)] + [c / (x-1)^2] =
a=2
b= -2
c= 3