Simplifying Polynomial Division Using Partial Fractions

[kasse]

(x^3+x^2+x-1)/(x^2+2x+2)

I want to compute the integral of this, but in order to use the method of partial fractions, I need to transform it into something with a higher degree in the divisor. How to I compute this?

It's supposed to be (x-1) + (x+1)/(x^2+2x+2)

Hope someone has got a clue

 

[courtrigrad]

$$\int \frac{x^{3}+x^{2}+x-1}{x^{2}+2x+2}$$Doing the long division you get $$x-1$$ with a remainder of $$\frac{x+1}{x^{2}+2x+2}$$

 

[radou]

(x^3+x^2+x-1)/(x^2+2x+2)

I want to compute the integral of this, but in order to use the method of partial fractions, I need to transform it into something with a higher degree in the divisor. How to I compute this?

It's supposed to be (x-1) + (x+1)/(x^2+2x+2)

Hope someone has got a clue

Not that I don't want to help, but polynomial division is pretty much of an 'algorythmic' procedure which can easily be found by google-ing.

 

[HallsofIvy]

Basically, it is just the same as long division of numbers- I'm sure you learned it long ago and just need a reminder.

Look at the leading terms: $x^2$ divides into $x^3$ x times. The first term of the quotient is x. Multiply the entire divisor by x: [/itex]x^3+ 2x^2+ 2x[/itex] and subtract: $x^3+ x^2+ x- 1- (x^3+ 2x^2- 2x)= -x^2- x- 1$. Now $x^2$ divides into $-x^2$ -1 times: quotient is now x-1. Multiply the divisor by -1 to get -x^2-2x-2 and subtract: -x^2- x- 1-(-x^2- 2x-2)= x+ 1. Since that has lower degree than the divisor you are done: the quotient is x- 1 with remainder x+1.
$$\frac{x^3+ x^2+ x- 1}{x^2+ 2x+ 2}= x- 1+ \frac{x+1}{x^2+ 2x+ 2}$$


NOW, do the hard part!

 

[kasse]

I found this algorithm at wikipedia [PLAIN]http://en.wikipedia.org/wiki/Polynomial_long_divisio[/URL]

The only thing I don't understand is why you don't pull down (-42) in step 3. -42-0 =-42...

 

[courtrigrad]

you can if you want to

 

[kasse]

OK, that makes sense.

Thank you all!