Partial Fraction `(x^2+15)/((x+3)^2(x^2+3))`
Solution:
1. Factors the denominator
`(x^2+15)/((x+3)^2(x^2+3))=(x^2+15)/((x+3)^2(x^2+3))`
2. Partial fraction for each factors
`:. (x^2+15)/((x+3)^2(x^2+3))=A/(x+3)+B/(x+3)^2+(Cx+D)/(x^2+3)`
3. Multiply through by the common denominator of `(x+3)^2(x^2+3)`
`:. x^2+15=A xx ((x+3)(x^2+3))+B xx (x^2+3)+(Cx+D) xx ((x+3)^2)`
`:. x^2+15=A xx (x^3+3x+3x^2+9)+B xx (x^2+3)+(Cx+D) xx (x^2+6x+9)`
`:. x^2+15=Ax^3+3Ax+3Ax^2+9A+Bx^2+3B+Cx^3+6Cx^2+9Cx+Dx^2+6Dx+9D`
4. Group the `x`-terms and the constant terms
`:. x^2+15=(A+C)x^3+(3A+B+6C+D)x^2+(3A+9C+6D)x+(9A+3B+9D)`
5. Coefficients of the two polynomials must be equal, so we get equations
`A+C=0`
`3A+B+6C+D=1`
`3A+9C+6D=0`
`9A+3B+9D=15`
Solution of equations using Elimination method
Total Equations are `4`
`a+0b+c+0d=0 -> (1)`
`3a+b+6c+d=1 -> (2)`
`3a+0b+9c+6d=0 -> (3)`
`9a+3b+0c+9d=15 -> (4)`
Select the equations `(1)` and `(2)`, and eliminate the variable `a`.
| `a+c=0` | ` xx 3->` | | `` | `3a` | | | `+` | `3c` | | | `=` | `0` | `` |
| | − | |
| `3a+b+6c+d=1` | ` xx 1->` | | `` | `3a` | `+` | `b` | `+` | `6c` | `+` | `d` | `=` | `1` | `` |
| | |
|
| | | | | `-` | `b` | `-` | `3c` | `-` | `d` | `=` | `-1` | ` -> (5)` |
Select the equations `(1)` and `(3)`, and eliminate the variable `a`.
| `a+c=0` | ` xx 3->` | | `` | `3a` | | | `+` | `3c` | | | `=` | `0` | `` |
| | − | |
| `3a+9c+6d=0` | ` xx 1->` | | `` | `3a` | | | `+` | `9c` | `+` | `6d` | `=` | `0` | `` |
| | |
|
| | | | | | | `-` | `6c` | `-` | `6d` | `=` | `0` | ` -> (6)` |
Select the equations `(1)` and `(4)`, and eliminate the variable `a`.
| `a+c=0` | ` xx 9->` | | `` | `9a` | | | `+` | `9c` | | | `=` | `0` | `` |
| | − | |
| `9a+3b+9d=15` | ` xx 1->` | | `` | `9a` | `+` | `3b` | | | `+` | `9d` | `=` | `15` | `` |
| | |
|
| | | | | `-` | `3b` | `+` | `9c` | `-` | `9d` | `=` | `-15` | ` -> (7)` |
Select the equations `(5)` and `(7)`, and eliminate the variable `b`.
| `-b-3c-d=-1` | ` xx 3->` | | | | `-` | `3b` | `-` | `9c` | `-` | `3d` | `=` | `-3` | `` |
| | − | |
| `-3b+9c-9d=-15` | ` xx 1->` | | | | `-` | `3b` | `+` | `9c` | `-` | `9d` | `=` | `-15` | `` |
| | |
|
| | | | | | | `-` | `18c` | `+` | `6d` | `=` | `12` | ` -> (8)` |
Select the equations `(6)` and `(8)`, and eliminate the variable `c`.
| `-6c-6d=0` | ` xx 3->` | | | | | | `-` | `18c` | `-` | `18d` | `=` | `0` | `` |
| | − | |
| `-18c+6d=12` | ` xx 1->` | | | | | | `-` | `18c` | `+` | `6d` | `=` | `12` | `` |
| | |
|
| | | | | | | | | `-` | `24d` | `=` | `-12` | ` -> (9)` |
Now use back substitution method
From (9)
`-24d=-12`
`=>d=(-12)/(-24)=1/2`
From (6)
`-6c-6d=0`
`=>-6c-6(1/2)=0`
`=>-6c-3=0`
`=>-6c=0+3=3`
`=>c=(3)/(-6)=-1/2`
From (5)
`-b-3c-d=-1`
`=>-b-3(-1/2)-(1/2)=-1`
`=>-b+1=-1`
`=>-b=-1-1=-2`
`=>b=2`
From (1)
`a+c=0`
`=>a+(-1/2)=0`
`=>a-1/2=0`
`=>a=0+1/2=1/2`
Solution using back substitution method.
`a=1/2,b=2,c=-1/2,d=1/2`
After solving these equations, we get
`a=1/2,b=2,c=-1/2,d=1/2`
Substitute these values in the original fraction
`((x^2+15))/((x+3)^2(x^2+3))=(1/2)/(x+3)+(2)/(x+3)^2+(-1/2x+1/2)/(x^2+3)`
This material is intended as a summary. Use your textbook for detail explanation.
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