T Timcago Junior Member Joined Apr 13, 2006 Messages 77 Apr 13, 2006 #1 F(x)= x^3-5x+7 G(x)= x^2-1 Do i factor x^2-1 and get (x+1)(x-1) and do the division separatly and add/subtract the to together at the end? How do you do the division on a problem like this?
F(x)= x^3-5x+7 G(x)= x^2-1 Do i factor x^2-1 and get (x+1)(x-1) and do the division separatly and add/subtract the to together at the end? How do you do the division on a problem like this?
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Apr 13, 2006 #2 Is this calculus? What are we to do with these two functions?
T Timcago Junior Member Joined Apr 13, 2006 Messages 77 Apr 13, 2006 #3 Sorry meant to put it on algebra I need to know how to divede F(x) by G(x)
skeeter Elite Member Joined Dec 15, 2005 Messages 3,204 Apr 13, 2006 #4 I assume youre dividing F(x) by G(x) ... \(\displaystyle \frac{F(x)}{G(x)}\) ever done long division? Code: x ---------------------------- x^2 - 1 | x^3 + 0x^2 - 5x + 7 x^3 - x --------------------------- -4x + 7 quotient is \(\displaystyle x + \frac{7 - 4x}{x^2 - 1}\)
I assume youre dividing F(x) by G(x) ... \(\displaystyle \frac{F(x)}{G(x)}\) ever done long division? Code: x ---------------------------- x^2 - 1 | x^3 + 0x^2 - 5x + 7 x^3 - x --------------------------- -4x + 7 quotient is \(\displaystyle x + \frac{7 - 4x}{x^2 - 1}\)
T Timcago Junior Member Joined Apr 13, 2006 Messages 77 Apr 13, 2006 #5 ya, i got that far. I was just thinking it was to easy to be true considering all my other problems were complex. The fact that x^2-1 could be factored through me off. Thank you.
ya, i got that far. I was just thinking it was to easy to be true considering all my other problems were complex. The fact that x^2-1 could be factored through me off. Thank you.