1. ## Algebraic division

I have this function:

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

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

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

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

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

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

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

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

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

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

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

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

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.