Extend $f(x)=\frac{6x^3-11x^2-16x-4}{2x^2-5x-2}$ in a way it is continuous $\forall x \in \mathbb{R}$

Question

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 ?

Answer 1

You can simply perform polynomial division to get

$$6x^3 -11x^2 -16x -4 = 3x(2x^2-5x-2) + 4x^2 - 10x-4 =\\ = 3x(2x^2-5x-2) + 2(2x^2 - 5x-2) = (3x+2)(2x^2 - 5x-2)$$

This explains the fact that the function has finite limits at points where the denominator $2x^2-5x-2$ is zero.

Thus, the desired extension is

$$f(x)=3x+2$$

Answer 2

Just do a polynomial long divison.

$$\frac{6x^3-11x^2-16x-4}{2x^2-5x-2} = 2+3x$$

Answer 3

One point here is that there is only a removable singularity in such a function if the numerator and denominator have a common root. Common factors of polynomials, which of course have identical roots, can be found by Euclid's extended division algorithm (here just dividing through gives what you need, but if there were a remainder the process could be iterated).

This process could be done entirely with polynomials having rational coefficients - no irrational numbers are required. At the end there may be a common factor which can cancel, and the remaining fraction will be in the equivalent of "lowest terms". At this stage, if the denominator still has a real root, each such root will be a singularity which cannot be removed.