Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$? I need to remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$, and find out what $f(7)$ equals. I am not sure what I've done wrong, but I'm getting 33, which the website I'm using tells me is wrong.
Please help me figure out what I'm doing wrong here.
My Steps:
$$\frac{x^2-11x+28}{x-7} \cdot \frac{x+7}{x+7}$$
$$=\frac{x^3+7x^2-11x^2-77x+28x+196}{x^2-49}$$
$$=\frac{x^3-4x^2-49x+196}{x^2-49}$$
$$=\frac{x^2(x-4)-49(x+4)}{x^2-49}$$
$$=(x-4)(x+4)$$
plug in $7$ for $x$.
$$(7-4)(7+4)=7^2-16=33$$

Hint: $x^2-11x+28=(x-7)(x-4)$. Can you finish? 
