Find position of roots

Let $f(x) = x^3+x^2-5x-1$ , then

$\begin{aligned} f'(x) & = 3x^2 + 2x - 5 & \small {\color{#3D99F6}\text{Putting }f'(x) = 0} \\ (3x+5)(x-1) & = 0 \\ f''(x) & = 6x + 2 \end{aligned}$

$f'(x) = 0$ , when

$\begin{cases} x = -\frac 53 & \implies f''\left(-\frac 53\right) = - 8 < 0 & \implies f\left(-\frac 53\right) = \frac {148}{27} & \small \text{ local maximum} \\ x = 1 & \implies f''(1) = 8 > 0 & \implies f(1) = -4 & \small \text{ local minimum} \end{cases}$

Let roots $a<b<c$ , then

• $a < - \frac 53 \approx - 1.667$ ; since $f(-3) = -4$ and $f(-2) = 5$ $\implies -3 < a < -2$ $\implies \lfloor a \rfloor = -3$
• $- 1.667 < b < 1$ ; since $f(-1) = 4$ and $f(0) = -1$ $\implies -1 < b < 0$ $\implies \lfloor b \rfloor = -1$
• $c > 1$ ; since $f(1) = -4$ and $f(2) = 1$ $\implies 1 < c < 2$ $\implies \lfloor c \rfloor = 1$

Therefore, $\lfloor a \rfloor + \lfloor b \rfloor + \lfloor c \rfloor = -3-1+1 = \boxed{-3}$