<s>[INST] Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from, 
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
 [/INST] Note that the denominator of $\frac{(x^2 + x-6)}{x-2}$ and 
$\frac{(x+3)(x-2)}{x-2}$ is the same, so it remains to show that the numerators are the same; i.e. that $x^2+x-6\equiv(x+3)(x-2)$.
Now, $\color{green}{(x+3)(x-2)} \equiv x^2\underbrace{-2x+3x}_{\equiv \ +x}-6 \equiv \color{green}{x^2+x-6},$ as required.
To expand brackets, use the distributive property (i.e. the fact that multiplication is distributive over addition).
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