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,
$$\frac{(x+3)(x-2)}{x-2}$$

Key fact: Knowing the roots of a polynomial (where the polynomial equals zero), let us factor it. 
So if $n$ and $m$ are two roots of the quadratic $ax^2+bx+c$, then we can factor it as $$ax^2+bx+c=a(x-n)(x-m).$$ The roots of a quadratic can be determined using the quadratic formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
To find out the roots of the quadratic $x^2+x-6$ use the above formula and you'll find out that they are: $-3$ and $2$. Therefore we can write our polynomial as: $$x^2+x-6=(x-(-3))(x-2)=(x+3)(x-2).$$
Hence, it follows that: $$\require{cancel}\frac{x^2 + x-6}{x-2} = \frac{(x+3)\color{red}{\cancel{\color{black}{(x-2)}}}}{\color{red}{\cancel{\color{black}{x-2}}}}=x+3.\tag{assuming $x\neq2$}$$
