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