Concerning Roots of the cubic equation $f(x)=x^3+x^2-5x-1$ and the Greatest Integer (or Floor) function

Question

The Question

I got into a rather tight corner with this question. It says:

Let $\alpha, \beta, \gamma$ be the roots of $f(x)=0$, where $f(x)=x^3+x^2-5x-1$. Then, the value of $\lfloor\alpha\rfloor+\lfloor\beta\rfloor+\lfloor\gamma\rfloor$ (where $\lfloor.\rfloor$ stands for the floor or greatest integer function) is:

1. $+1$
2. $-2$
3. $+4$
4. $-3$

What I've tried

If we take $f(x)=x^3+x^2-5x-1$ and compare it with the generalized form $$f(x)=ax^3+bx^2+cx+d$$ we see that: $a=1, b=1, c=-5$ and $d=-1$.

Now, the following formulas are known:

1. $$\alpha+\beta+\gamma=\frac{-b}{a}=-1\tag{1}$$
2. $$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}=-5\tag{2}$$
3. $$\alpha\beta\gamma=\frac{-d}{a}=1\tag{3}$$

But I don't know what to do next. I tried putting in values of $x$ from $-2$ to $2$ to see it $f(x)$ equals $0$ for any particular value, but without any luck. I know that in $(1)$, $\alpha+\beta+\gamma=-1$. But that, apparently, seems useless as the answer is (surprisingly):

Option 4., i.e. $-3$

How can I work my way out of this? There's no need for a full answer (unless you absolutely insist); I could do with a hint.

Answer 1

Using Descartes' rule, we can solve it as follows :

$f(2)>0$ & $f(1)<0 $. Thus, $\lfloor{\alpha}\rfloor=1$

$f(-1)>0$ & $f(0)<0 $. Thus, $\lfloor{\beta}\rfloor=-1$

$f(-3)<0$ & $f(-2)>0 $. Thus, $\lfloor{\gamma}\rfloor=-3$

Thus, $\lfloor{\alpha}\rfloor + \lfloor{\beta}\rfloor + \lfloor{\gamma}\rfloor = -3$

Answer 2

This one will do even if you are unaware of the Descartes' rule:

Draw a rough graph of the function and observe that the graph cuts the x- axis thrice: between $1$ and $2$, between $-1$ and $0$ and once between $-3$ and $-2$. You can reach these conclusions by calculating the value of $f$ at these points.

Thus, we have roots between these points and by taking their greatest integer function values, the required answer is $-3-1+1=-3$