6. Example `(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)`
Partial Fraction `(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)`
Solution:
The numerator `x^5-2x^4+x^3+x+5` is of degree 5 and the denominator `x^3-2x^2+x-2` is of degree 3. So first do the long division
long division of `x^5-2x^4+x^3+x+5` and `x^3-2x^2+x-2`Final Solution | | `` | `x^2` | | | | | | | | | | | | | | `color{blue}{x^3-2x^2+x-2}` | `` | `x^5` | `-` | `2x^4` | `+` | `x^3` | `+` | `0x^2` | `+` | `x` | `+` | `5` | | | | | `` | −`x^5` | `-` | +`2x^4` | `+` | −`x^3` | `-` | +`2x^2` | | | | | | `color{green}{x^2} xx (color{blue}{x^3-2x^2+x-2})` | | | | | | | | | `` | `2x^2` | `+` | `x` | `+` | `5` | | |
Final answer `= "Quotient" + (color{Magenta}{"Remainder"})/(color{blue}{"Divisor"})`. `:.` Final answer = `x^2 + (color{Magenta}{2x^2+x+5})/(color{blue}{x^3-2x^2+x-2})` Here, Divisor = `x^3-2x^2+x-2` Dividend = `x^5-2x^4+x^3+x+5` Quotient = `x^2` Remainder = `2x^2+x+5`
Step by step division solutionStep - 1 : 1. Divide the first term of the dividend by the first term of the divisor : `(x^5)/(x^3)=color{green}{x^2}` 2. Write down the calculated result `color{green}{x^2}` in the upper part of the table. 3. Multiply it by the divisor `color{green}{x^2} xx (color{blue}{x^3-2x^2+x-2})=color{red}{x^5-2x^4+x^3-2x^2}` 4. Subtract this result from the dividend `(x^5-2x^4+x^3+0x^2+x+5)-(color{red}{x^5-2x^4+x^3-2x^2})=color{Magenta}{2x^2+x+5}` | | `` | `x^2` | | | | | | | | | | | | | | `color{blue}{x^3-2x^2+x-2}` | `` | `x^5` | `-` | `2x^4` | `+` | `x^3` | `+` | `0x^2` | `+` | `x` | `+` | `5` | | | | | `` | −`x^5` | `-` | +`2x^4` | `+` | −`x^3` | `-` | +`2x^2` | | | | | | `color{green}{x^2} xx (color{blue}{x^3-2x^2+x-2})` | | | | | | | | | `` | `2x^2` | `+` | `x` | `+` | `5` | | |
The long division gives
`(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2) = x^2 + (2x^2+x+5)/(x^3-2x^2+x-2)`
Now find partial fraction of `(2x^2+x+5)/(x^3-2x^2+x-2)`
1. Factors the denominator
`(2x^2+x+5)/(x^3-2x^2+x-2)=(2x^2+x+5)/((x-2)(x^2+1))`
2. Partial fraction for each factors
`:. (2x^2+x+5)/((x-2)(x^2+1))=A/(x-2)+(Bx+C)/(x^2+1)`
3. Multiply through by the common denominator of `(x-2)(x^2+1)`
`:. 2x^2+x+5=A xx (x^2+1)+(Bx+C) xx (x-2)`
`:. 2x^2+x+5=Ax^2+A+Bx^2-2Bx+Cx-2C`
4. Group the `x`-terms and the constant terms
`:. 2x^2+x+5=(A+B)x^2+(-2B+C)x+(A-2C)`
5. Coefficients of the two polynomials must be equal, so we get equations
`A+B=2`
`-2B+C=1`
`A-2C=5`
Solution of equations using Elimination methodTotal Equations are `3` `a+b+0c=2 -> (1)` `0a-2b+c=1 -> (2)` `a+0b-2c=5 -> (3)`
Select the equations `(1)` and `(3)`, and eliminate the variable `a`. | `a+b=2` | ` xx 1->` | | `` | `a` | `+` | `b` | | | `=` | `2` | `` | | | − | | | `a-2c=5` | ` xx 1->` | | `` | `a` | | | `-` | `2c` | `=` | `5` | `` | | | |
| | | | | | `` | `b` | `+` | `2c` | `=` | `-3` | ` -> (4)` |
Select the equations `(2)` and `(4)`, and eliminate the variable `b`. | `-2b+c=1` | ` xx -1->` | | | | `` | `2b` | `-` | `c` | `=` | `-1` | `` | | | − | | | `b+2c=-3` | ` xx 2->` | | | | `` | `2b` | `+` | `4c` | `=` | `-6` | `` | | | |
| | | | | | | | `-` | `5c` | `=` | `5` | ` -> (5)` |
Now use back substitution method From (5) `-5c=5` `=>c=(5)/(-5)=-1` From (2) `-2b+c=1` `=>-2b+(-1)=1` `=>-2b-1=1` `=>-2b=1+1=2` `=>b=(2)/(-2)=-1` From (1) `a+b=2` `=>a+(-1)=2` `=>a-1=2` `=>a=2+1=3` Solution using back substitution method. `a=3,b=-1,c=-1`
After solving these equations, we get
`a=3,b=-1,c=-1`
Substitute these values in the original fraction
`((x^5-2x^4+x^3+x+5))/((x-2)(x^2+1))=x^2+(3)/(x-2)+(-1x-1)/(x^2+1)`
This material is intended as a summary. Use your textbook for detail explanation.
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