x^2

____

3x + 1

How do I simplify this to get the Slant Asymptote?

I thought that you couldn't divide x^2 by 3x +1 because their degrees don't match.

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- Oct 5th 2009, 02:57 PMIsimbotPolynomial Long Division for Rational Functions
x^2

____

3x + 1

How do I simplify this to get the Slant Asymptote?

I thought that you couldn't divide x^2 by 3x +1 because their degrees don't match. - Oct 5th 2009, 03:30 PMIsimbot
Oh, wait!

Is it 1/3x ? - Oct 6th 2009, 04:32 AMHallsofIvy
Is

**what**"1/3x"? And do you mean 1/(3x) or (1/3)x?

To divide $\displaystyle x^2$ by 3x+1, first write [tex]x^2= x^2+ 0x+ 0.

How many times does "3x" divide into $\displaystyle x^2$? (1/3)x times, of course. But you are not done. Now subtract 3x+1 times (1/3)x from $\displaystyle x^2$: $\displaystyle x^2- (1/3)x(3x+1)= x^2- x^2- (1/3)x= -(1/3)x+ 0$. How many times does 3x divide into -(1/3)x? -1/9 times. And now you subtract (3x+1) times -1/9 from -(1/3)x+ 0: -(1/3)x- (-1/9)(3x+1)= -(1/3)x+ (1/3)x+ 1/9= 1/9.

3x+1 divides into $\displaystyle x^2$ (1/3)x- 1/9 times with remainder 1/9.