1. ## Algebraic division

I have this function:

$\displaystyle f(x) = x + \frac{3}{x - 1} - \frac{12}{x^2 + 2x - 3}$

We've been asked to show that the above equation can be written like this:

$\displaystyle \frac{x^2 + 3x + 3}{x + 3}$

I just can't seem to get it and i can't figure out why!

Could someone show me how it's done please?

Many thanks.

2. $\displaystyle x+\frac{3}{x-1}-\frac{12}{x^2+2x-3}$

$\displaystyle =x+\frac{3}{x-1}-\frac{12}{(x-1)(x+3)}$

$\displaystyle =\frac{x(x-1)(x+3)+3(x+3)-12}{(x-1)(x+3)}$

$\displaystyle =\frac{x^3+2x^2-3x+3x+9-12}{(x-1)(x+3)}$

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

For $\displaystyle x^3+2x^2-3$, test 1:
$\displaystyle (1)^3+2(1)^2-3=0$

Hence, $\displaystyle x-1$ is a factor of $\displaystyle x^3+2x^2-3$. Using polynomial long division, you get the other factor to be $\displaystyle x^2+3x+3$.

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

$\displaystyle =\frac{x^2+3x+3}{x+3}$

3. Thanks a lot! I've been looking at that for days and got as far as where you found x -1 to be a factor.