Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$?

Question

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 $x$.

$$(7-4)(7+4)=7^2-16=33$$

Answer 1

To explain why you get the wrong answer, in your second last line, the last term of the numerator is $-49(x-4)$ and not $-49(x+4).$

Faster approach:

$$\frac{x^2-11x+28}{x-7}=\frac{(x-7)(x-4)}{x-7}=x-4$$

Hence to remove the discontinuity at $7$, the right value to redefine at $f(7)$ should be $7-4$.

Factorization should be the way to go, multiplying by the conjugate doesn't serve the purpose here.

Answer 2

x=7 is a root of $x^2-11x+28$. So write $x^2-11x+28=(x-7)p(x)$. Then $f(7)=p(7)$

Answer 3

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

Answer 4

You probably want to remove it at $7$ and not at $f(7)$. Write $$ f(x) = {(x-7)(x-4)\over x-7 } = x-4$$

So $$\lim_{x\to 7} f(x) =3$$