Finding roots of cubic equation If $\alpha,\beta,\gamma $ are the roots of the equation $2x^3-3x^2-12x+1=0$.Then find the value of [$\alpha$]+[$\beta$]+[$\gamma$],where [.] denotes greatest integer function.
My attempt:
I first tried hit and trial to guess the first root but no luck.Then I tried rational root method, but its roots are not rational (both roots could not satisfy the equation.)
I think its roots are irrational. Can someone help me finding its roots?
 A: Localise the roots between two consecutive integers and use the Intermediate value theorem.
Let $p(x)=2x^3-3x^2-12x+1$. As $p'(x)=6x^2-6x-12=6(x+1)(x-2)$, $p(x)$  has a local maximum at $x=-1$ and a local minimum at $x=2$:
$$p(-1)=8,\quad p(2)=-19.$$
Thus we know there are three real roots: $x1<-1$, $-1< x_2 <2$, $x_3>2$.  Now a table of the values of $p(x)$ at some integer points will do:
$$\begin{matrix}x&p(x)\\\hline-2&-3\\-1&8\\0&1\\1&-12\\2&-19\\3&-8\\4&33\end{matrix}$$
Hence $\lfloor x_1\rfloor=-2$, $\lfloor x_2\rfloor=0$, $\lfloor x_3\rfloor=3$, so that $\lfloor x_1\rfloor+\lfloor x_2\rfloor+\lfloor x_3\rfloor=1$.
A: Let $$f(x)=2x^3-3x^2-12x+1$$  
Now, 
$f(0)>0$ and $f(1)<0 \implies 0< \alpha <1$  
$f(-2)<0$ and $f(-1)>0 \implies -2< \beta <-1$  
$f(3)<0$ and $f(4)>0 \implies 3< \gamma <4$   
$\therefore \lfloor\alpha\rfloor + \lfloor\beta\rfloor + \lfloor\gamma\rfloor = 0+(-2)+3=1$
A: Given $f(x)=2x^3-3x^2-12x+1$, we have a sign change between $-2$ and $-1$, between $0$ and $1$, between $3$ and $4$.
