Find values of $a$ and $b$ $\in \mathbb{R}$ make the function continuous everywhere. 
I need some help for this question, I’ve tried to use the definition of what it means to be continuous by solving the left limit and the right limit of 2. But it doesn't really give me any useful information.
Would really appreciate for any help.Thanks in advance.
 A: For $x\neq 2$ there are not problems , the function $f$ si continuos. 
As regards the Point $x=2$, you must impose the condition of continuity:
$\lim_{x\to 2}f(x)=f(2)=b$
but 
$\lim_{x\to 2}f(x)$ is finite if and only if 
$(x-2)|f(x)$ 
so you have 
$f(x)=(x-2)(dx^2+ex+f)=dx^3+ex^2+fx -2dx^2-2ex-2f$
$=3x^3+ax^2+x-6$
that means 
$d=3$, $e-2d=a$, $f-2e=1$ and $-2f=-6$.
Thus 
$d=3$,  $e=1$, $f=3$ and $a=e-2d=1-6=-5$
So you must impose $a=-5$ and you have that
$f(x)=(x-2)(3x^2+x+3)$ 
The condition of continuity tells su that
$32^2+2+3=b$
So 
$a=-5$ and $b=17$
are the unique solutions for which $f$ results to be continuos on $x=2$
A: First of all the function is continuous for every $x\neq 2$. 
Now we have to make $f$ continuous at $2$.
Lets look at a simple example:
$g:\mathbb{R}\to\mathbb{R}$, $g(x)=\begin{cases}2x+1, x\geq 4\\ b, x<4\end{cases}$
Now we have to find $b$ such that $g$ is continuous for every $x\in\mathbb{R}$.
For that we would just take $g(4)=9=b$. Then $g$ is continuous at x=4 (the only critical point).
For your example you have to do basically the same. It just involves more calculation.
We have to find $a$ too. 
You have to find a such that (x-2) is a linear factor of $3x^3+ax^2+x-6$.
So for $h(x)=3x^3+ax^2+x-6$ it has to be $h(2)=0$. This gives $a$.
You can then find $b$ as in the example above.
A: Your function is clearly continuous everywhere except perhaps $x=2$. What you need is that $f$ satisfies that:
$$ \underset{ x\rightarrow 2 }{\lim } f(x) \; \text{exists} \quad \text{and} \quad \underset{ x\rightarrow 2 }{\lim } f(x)=b $$
For the first part to hold you need that $x=2$, is also a root of the numenator. Otherwise the limit will diverge. This will give you some condition on $a$, and the ability to write $f$ of the form $f(x)=\alpha x^2+ \beta x+ \gamma$ for all $x\neq 2$. $b$ has to satisfy that $\alpha x^2+ \beta x+ \gamma \overset{x\rightarrow 2}{\rightarrow}b$.
These are the restrictions you'll get. Any valid $a$ and $b$ for which this works will be okay.
