resolve the partial fraction:x^3+6/x^2(x+3).with explanations thanks
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))
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!
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.
tank you reiny
To resolve the given partial fraction `(x^3 + 6) / (x^2(x + 3))`, we will follow a step-by-step approach:
Step 1: Factorization
First, let's factor the numerator and denominator:
Numerator: `x^3 + 6`
Denominator: `x^2(x + 3)`
No further factorization is possible for the given numerator, but we can factor the denominator further by splitting it into two separate factors: `x^2` and `(x + 3)`.
Step 2: Form the Partial Fraction Decomposition
Since the degree of the numerator is equal to or greater than the degree of the denominator, we need to use polynomial long division to divide the numerator by the denominator.
We have:
```
------- A
x^2 | x^3 + 0x^2 + 0x + 6
- (x^3 + 3x^2)
- 3x^2 + 0x + 6
```
Now, we will factor the remaining polynomial `(-3x^2 + 0x + 6)`.
```
------- A B
x^2 | x^3 + 0x^2 + 0x + 6
- (x^3 + 3x^2)
- 3x^2 + 0x + 6
= 0x^3 - 3x^2 + 0x + 6
```
Step 3: Equating Coefficients
Equating the coefficients of the corresponding powers of `x`, we get the following equations:
For `x^2` terms:
`0A + B = -3`
For constant terms:
`6A = 6`
Simplifying these equations, we find that `A = 1` and `B = -3`.
Step 4: Writing the Partial Fraction Decomposition
Now, we can rewrite the initial fraction in terms of partial fractions:
`(x^3 + 6) / (x^2(x + 3)) = A / x^2 + B / (x + 3)`
Substituting the values of `A` and `B` that we found earlier, we have:
`(x^3 + 6) / (x^2(x + 3)) = 1 / x^2 - 3 / (x + 3)`
Therefore, the resolved partial fraction is:
`(x^3 + 6) / (x^2(x + 3)) = 1 / x^2 - 3 / (x + 3)`