Partial Fraction `(x-3)/(x^3+2x^2+x)`
Solution:
1. Factors the denominator
`(x-3)/(x^3+2x^2+x)=(x-3)/(x(x+1)^2)`
2. Partial fraction for each factors
`:. (x-3)/(x(x+1)^2)=A/(x)+B/(x+1)+C/(x+1)^2`
3. Multiply through by the common denominator of `x(x+1)^2`
`:. x-3=A xx ((x+1)^2)+B xx (x(x+1))+C xx (x)`
`:. x-3=A xx (x^2+2x+1)+B xx (x^2+x)+C xx (x)`
`:. x-3=Ax^2+2Ax+A+Bx^2+Bx+Cx`
4. Group the `x`-terms and the constant terms
`:. x-3=(A+B)x^2+(2A+B+C)x+A`
5. Coefficients of the two polynomials must be equal, so we get equations
`A+B=0`
`2A+B+C=1`
`A=-3`
Solution of equations using Elimination method
Total Equations are `3`
`a+b+0c=0 -> (1)`
`2a+b+c=1 -> (2)`
`a+0b+0c=-3 -> (3)`
Select the equations `(1)` and `(2)`, and eliminate the variable `a`.
| `a+b=0` | ` xx 2->` | | `` | `2a` | `+` | `2b` | | | `=` | `0` | `` |
| | − | |
| `2a+b+c=1` | ` xx 1->` | | `` | `2a` | `+` | `b` | `+` | `c` | `=` | `1` | `` |
| | |
|
| | | | | `` | `b` | `-` | `c` | `=` | `-1` | ` -> (4)` |
Select the equations `(1)` and `(3)`, and eliminate the variable `a`.
| `a+b=0` | ` xx 1->` | | `` | `a` | `+` | `b` | | | `=` | `0` | `` |
| | − | |
| `a=-3` | ` xx 1->` | | `` | `a` | | | | | `=` | `-3` | `` |
| | |
|
| | | | | `` | `b` | | | `=` | `3` | ` -> (5)` |
Select the equations `(4)` and `(5)`, and eliminate the variable `b`.
| `b-c=-1` | ` xx 1->` | | | | `` | `b` | `-` | `c` | `=` | `-1` | `` |
| | − | |
| `b=3` | ` xx 1->` | | | | `` | `b` | | | `=` | `3` | `` |
| | |
|
| | | | | | | `-` | `c` | `=` | `-4` | ` -> (6)` |
Now use back substitution method
From (6)
`-c=-4`
`=>c=4`
From (4)
`b-c=-1`
`=>b-(4)=-1`
`=>b-4=-1`
`=>b=-1+4=3`
From (1)
`a+b=0`
`=>a+(3)=0`
`=>a+3=0`
`=>a=0-3=-3`
Solution using back substitution method.
`a=-3,b=3,c=4`
After solving these equations, we get
`a=-3,b=3,c=4`
Substitute these values in the original fraction
`((x-3))/(x(x+1)^2)=(-3)/(x)+(3)/(x+1)+(4)/(x+1)^2`
This material is intended as a summary. Use your textbook for detail explanation.
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