Partial Fraction `(2x^3)/((x+1)(x-1))`Solution:The numerator `2x^3` is of degree 3 and the denominator `x^2-1` is of degree 2. So first do the long division
long division of `2x^3` and `x^2-1`
Final Solution
| | `` | `2x` | `+` | `0` | | | | | | |
| `color{blue}{x^2-1}` | `` | `2x^3` | `+` | `0x^2` | `+` | `0x` | `+` | `0` | | |
| | `` | −`2x^3` | | | `-` | +`2x` | | | | `2x xx (color{blue}{x^2-1})` |
| | | | `` | `0x^2` | `+` | `2x` | `+` | `0` | | |
| | | | `` | −`0x^2` | | | `-` | +`0` | | `color{green}{0} xx (color{blue}{x^2-1})` |
| | | | | | `` | `2x` | | | | |
Final answer `= "Quotient" + (color{Magenta}{"Remainder"})/(color{blue}{"Divisor"})`.
`:.` Final answer = `2x+0 + (color{Magenta}{2x})/(color{blue}{x^2-1})`
Here, Divisor = `x^2-1`
Dividend = `2x^3`
Quotient = `2x+0`
Remainder = `2x`
Step by step division solution
Step - 1 : 1. Divide the first term of the dividend by the first term of the divisor : `(2x^3)/(x^2)=color{green}{2x}`
2. Write down the calculated result `color{green}{2x}` in the upper part of the table.
3. Multiply it by the divisor `color{green}{2x} xx (color{blue}{x^2-1})=color{red}{2x^3-2x}`
4. Subtract this result from the dividend
`(2x^3+0x^2+0x+0)-(color{red}{2x^3-2x})=color{Magenta}{0x^2+2x+0}`
| | `` | `2x` | | | | | | | | |
| `color{blue}{x^2-1}` | `` | `2x^3` | `+` | `0x^2` | `+` | `0x` | `+` | `0` | | |
| | `` | −`2x^3` | | | `-` | +`2x` | | | | `color{green}{2x} xx (color{blue}{x^2-1})` |
| | | | `` | `0x^2` | `+` | `2x` | `+` | `0` | | |
Step - 2 : 1. Divide the first term of the dividend by the first term of the divisor : `(0x^2)/(x^2)=color{green}{0}`
2. Write down the calculated result `color{green}{0}` in the upper part of the table.
3. Multiply it by the divisor `color{green}{0} xx (color{blue}{x^2-1})=color{red}{0x^2-0}`
4. Subtract this result from the remainder
`(0x^2+2x+0)-(color{red}{0x^2-0})=color{Magenta}{2x}`
| | `` | `2x` | `+` | `0` | | | | | | |
| `color{blue}{x^2-1}` | `` | `2x^3` | `+` | `0x^2` | `+` | `0x` | `+` | `0` | | |
| | `` | −`2x^3` | | | `-` | +`2x` | | | | `2x xx (color{blue}{x^2-1})` |
| | | | `` | `0x^2` | `+` | `2x` | `+` | `0` | | |
| | | | `` | −`0x^2` | | | `-` | +`0` | | `color{green}{0} xx (color{blue}{x^2-1})` |
| | | | | | `` | `2x` | | | | |
The long division gives
`(2x^3)/(x^2-1) = 2x + (2x)/(x^2-1)`
Now find partial fraction of `(2x)/(x^2-1)`
1. Factors the denominator
`(2x)/(x^2-1)=(2x)/((x-1)(x+1))`
2. Partial fraction for each factors
`:. (2x)/((x-1)(x+1))=A/(x-1)+B/(x+1)`
3. Multiply through by the common denominator of `(x-1)(x+1)`
`:. 2x=A xx (x+1)+B xx (x-1)`
`:. 2x=Ax+A+Bx-B`
4. Group the `x`-terms and the constant terms
`:. 2x=(A+B)x+(A-B)`
5. Coefficients of the two polynomials must be equal, so we get equations
`A+B=2`
`A-B=0`
Solution of equations using Elimination method
Total Equations are `2`
`a+b=2 -> (1)`
`a-b=0 -> (2)`
Select the equations `(1)` and `(2)`, and eliminate the variable `a`.
| `a+b=2` | ` xx 1->` | | `` | `a` | `+` | `b` | `=` | `2` | `` |
| | − | |
| `a-b=0` | ` xx 1->` | | `` | `a` | `-` | `b` | `=` | `0` | `` |
| | |
|
| | | | | `` | `2b` | `=` | `2` | ` -> (3)` |
Now use back substitution method
From (3)
`2b=2`
`=>b=(2)/(2)=1`
From (1)
`a+b=2`
`=>a+(1)=2`
`=>a+1=2`
`=>a=2-1=1`
Solution using back substitution method.
`a=1,b=1`
After solving these equations, we get
`a=1,b=1`
Substitute these values in the original fraction
`((2x^3))/((x-1)(x+1))=2x+(1)/(x-1)+(1)/(x+1)`
This material is intended as a summary. Use your textbook for detail explanation.
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