after breaking down this fraction [math]\frac{x^2+2}{x^3-x}[/math]

I got [math]\frac{-2}{x} + \frac{3}{2(x-1)} + \frac{1}{2(x+1)}[/math]

but using my ti-89 and mathematic there is an extra +1 at the end so I was wondering where does the +1 come from?

mathematic and ti-89 gave [math]\frac{-2}{x}+\frac{3}{2(x-1)}+\frac{1}{2(x+1)}+1[/math]

after breaking down this fraction (x^2+2)/(x^3-x)

I got (-2)/(x)+3/(2(x-1))+1/(2(x+1))

but using my ti-89 and mathematic there is an extra +1 at the end so I was wondering where does the +1 come from?

mathematic and ti-89 gave (-2)/(x)+3/(2(x-1))+1/(2(x+1))+1

 

Try to use LaTex where possible' date=' makes things easier to read

 

after breaking down this fraction [math]\frac{x^2+2}{x^3-x}[/math]

I got [math]\frac{-2}{x} + \frac{3}{2(x-1)} + \frac{1}{2(x+1)}[/math]

but using my ti-89 and mathematic there is an extra +1 at the end so I was wondering where does the +1 come from?

mathematic and ti-89 gave [math]\frac{-2}{x} + \frac{3}{2(x-1)} + \frac{1}{2(x+1)} + 1[/math]

 

 

... is that what you meant?

 

Cheers,

 

Ryan Jones

yep, thanks, I have no clue how to use LaTex. I am noob...

yep, thanks, I have no clue how to use LaTex. I am noob...

 

I understand

 

If you need help you can try looking at the examples here and you can ask the questions there too and I'm shure someone will be glad to help you

 

Cheers,

 

Ryan Jones

click on any LaTeX image to view the code

ooo, I know how to latex now. heh

so back to the question:)

so I was wondering where does the +1 come from?

 

It comes from you entering the expression into your calculator wrong. You put [imath]x^3+2[/imath] in for the numerator.

woops, I typed it wrong here I meant [math]\frac{x^3+2}{x^3-x}[/math] so why is there extra +1 for mathematica and calculator? when I do it I get [math] \frac{A}{X} + \frac{B}{X-1} + \frac{C}{X+1}[/math] that is how I got my 3 terms.

If you do a long division first it is obvious

[math]

\frac{(x^3-x)+(x +2)}{x^3-x} = 1 + \frac{x+2}{x^3-x}

[/math]

thank you, I didn't see how the numerator is bigger than the denominator, so when the largest power is same on top and bottom, long division is neccesary? This one looked like a proper fraction to me

Yes, if you have a fraction with the polynominal on the numerator greater than or equal to the degree of the polynominal in the demoninator, then you will need to perform long division.