Question

resolve the partial fraction:x^3+6/x^2(x+3).with explanations thanks

Answer 1

First, since the degree of the top is at least as big as that of the bottom, just do a long division, and you get

(x^3+6)/(x^2 (x+3)) = 1 + (-3x^2+6)/(x^2 (x+3))

Now you want the partial fractions for the remainder, in the form

(-3x^2+6)/(x^2 (x+3)) = A/x + B/x^2 + C/(x+3)
=
Ax(x+3)+B(x+3)+Cx^2
------------------------
x^2(x+3)
=
Ax^2+3Ax+Bx+3B+Cx^2
----------------------
x^2(x+3)
=
(A+C)x^2 + (3A+B)x + 3B
-------------------------
x^2(x+3)

In order for those two fractions to be identical, all the coefficients of all the powers of x must match. That means we have

A+C = -3
3A+B = 0
3B = 6

That's easy to solve, and we get
A = -2/3
B=2
C = -7/3

so our partial fraction for the complete original fraction is

1 + (-2/3)/x + 2/x^2 + (-7/3)/(x+3)

or 1 - 2/(3x) + 2/x^2 - 7/(3(x+3))

Answer 2

You fixed Damon's concern, but still brackets are missing
I will assume you meant:

(x^3+6)/(x^2(x+3) )

I first did a long algebraic division to get it to
1 - (3x^2 - 6)/(x^2(x+3) )

So let's just work on that last term

let (3x^2 - 6)/(x^2(x+3) ) = A/x + (Bx + C)/x^2 + D/(x+3)
= ( Ax(x+3) + (x+3)(Bx + C) + Dx^2)/(x^2(x+3))

then 3x^2 - 6 = Ax(x+3) + (x+3)(Bx + C) + Dx^2
let x = 0 , -6 = 0 + 3C + 0
C = -2
let x = -3, 21 = 0 + 0 + 9D
D = 21/9 = 7/3
let x = 1 , -3 = 4A + 4(B - 2) + D = -3
4A + 4B - 8 + 7/3 = -3
A + B = 2/3

let x = 3 --> 18A + 6(3B -2) + 9D = 21
18A + 18B -12 + 63/3 = 21
18A + 18B = 12
A+B = 2/3

no matter what I try for x, I get A + B = 2/3
So we can let A and B be anything, why not make it simple and let B = 0, then A = 2/3

then :
(3x^2 - 6)/(x^2(x+3) ) = (2/3) / x + (-2)/x^2 + (7/3) / (x+3)

= 2/(3x) - 2/x^2 + 7/(3(x+3))

finally:

(x^3+6)/(x^2(x+3) ) = 1 - (3x^2 - 6)/(x^2(x+3))
= 1 - (2/(3x) - 2/x^2 + 7/(3(x+3)) )

= 1 - 2/(3x) + 2/x^2 - 7/(3(x+3))

Wheuhhh!

Answer 3

anonymous made the "lucky" assumption that the second fraction ?/x^2 had a constant for a numerator.
In general you cannot make such an assumption.
If the denominator is a second degree expression, like x^2, then the numerator could be a first degree expression of the form Bx + k

notice the values of A,B, C, and D are not defined the same way as those by anonymous.

Answer 4

tank you reiny