Roots of a Rational Function

We have $f(x) = \frac {x^2 - 13x + 22}{x^2 - 8x + 12}$ we can factorise it to : $f(x) = \frac {(x-2) (x-11)}{(x-2) (x-6)}$

now we get the root candidates of $f(x)$ are 2,6, and 11. Since substituting 2 or 6 as the $x$ will lead to zero on the denominator (which is undefined result), the only valid root is 11. So the answer is 11

we have $x^2 - 13x +22 = (x-11)(x-2)$ and $x^2 -8x+12 = (x-2)(x-6)$ therefore the only root of f(x) is 11

If we factorize: $x^2-13x+22=(x-11)·(x-2)$ and $x^2-8x+12=(x-2)·(x-6)$ Then, $\frac{(x-11)·(x-2)}{(x-2)·(x-6)}=\frac{x-11}{x-6}$ So the only root is x=11 and thus the sum of all roots is S=11

f(x) becomes 0 at x=11 and x=2 but we cannot count x=2 as a root because x=2 is also a root of denominator equation which makes it undefined.Hence,we have to neglect x=2 and the final answer would be x=11

First I factored the bottom to figure out what it could not be. Then I factored the top to figure out what it could be (looking for zeros in both). The bottom had 6 and 2, so it couldn't be those. The top had 11 and 2. It couldn't be 2, so the answer was 11

11