<s>[INST] Extend $f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$ in a way it is continuous $\forall x \in \mathbb{R}$  Problem 
If it is possible to extend $$f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$$ in a way it is continuous $\forall x \in \mathbb{R}$. Extend all $x \not\in X$ where $X$ is domain of $f$. Also define domain $X$.
 Attempt to solve 
By solving $2x^2-5x-2=0$ we can get $\forall x \not\in X$.
$$ x= \frac{5\pm \sqrt{(-5)^2-4 \cdot 2\cdot(-2)}}{2 \cdot 2} $$
$$ x= \frac{5 \pm \sqrt{41}}{4} $$
Meaning :
$$ X=\mathbb{R}\setminus \{\frac{5-\sqrt{41}}{4},\frac{5+\sqrt{41}}{4}\} $$
In order to $f$ to be continuous $\forall x \
\not\in X$ have to have limits in these undefined points.
$$ \lim_{x \rightarrow \frac{5-\sqrt{41}}{4}} \frac{6x^3-11x^2-16x-4}{2x^2-5x-2} 
 =\frac{1}{4}(23-3\sqrt{41})$$
and 
$$ \lim_{x \rightarrow \frac{5+\sqrt{41}}{4}}\frac{6x^3-11x^2-16x-4}{2x^2-5x-2} = \frac{1}{4}(23+3\sqrt{41})$$
Meaning if i set 
$$ 
\begin{cases}
f(\frac{5-\sqrt{41}}{4})=\frac{1}{4}(23-3\sqrt{41}) \\
f(\frac{5+\sqrt{41}}{4}=\frac{1}{4}(23+3\sqrt{41})
\end{cases}
$$
is continous $\forall x \in \mathbb{R}$
Now only problem is that is this correct solution and if it is how do i solve it more simply than this ? I didn't compute these limits by hand, maybe there is more simpler solution ?
 [/INST] Just do a polynomial long divison.
$$\frac{6x^3-11x^2-16x-4}{2x^2-5x-2} = 2+3x$$
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