Hidden Roots ........

On drawing the graph of the polynomial it will intersect $x$ axis 3 times. Between -1 and 0 , b/w 2 and 3 and b/w 8 and 9 so simply by the answer would become -1 +2+8 =9.

Let's use Cardano's method:

Make the substitution $x=\dfrac{y+11}{3}$ to delete the quadratic term, and multiply by $27$ :

$\left(\dfrac{y+11}{3}\right)^3-11\left(\dfrac{y+11}{3}\right)^2+19\left(\dfrac{y+11}{3}\right)+13=0 \\ y^3-192y-430=0$

Now, compare that equation with the identity $(u+v)^3-3uv(u+v)-(u^3+v^3)=0$ . An equation system will be formed:

$y=u+v \\ 3uv=192 \Longrightarrow u^3v^3=262144 \\ u^3+v^3=430$

Make an equation in $z$ with roots $u^3$ and $v^3$ :

$(z-u^3)(z-v^3)=z^2-(u^3+v^3)z+u^3v^3=0 \\ z^2-430z+262144=0$

Using the quadratic formula we get:

$z=215 \pm 3\sqrt{23991}i$

Now, let's find the absolute value and the argument of $z$ :

$|z|=\sqrt{215^2+(3\sqrt{23991})^2} \\ |z|=512$

$\arg(z)=sgn(Im(z))\arccos\left(\dfrac{Re(z)}{|z|}\right) \\ \arg(z)=\pm \arccos\left(\dfrac{215}{512}\right)$

So,

$u=\sqrt[3]{|z|} cis\left(\dfrac{\arg(z)}{3}\right)$ , $v=\sqrt[3]{|z|} cis\left(-\dfrac{\arg(z)}{3}\right)$

$y=u+v=2\sqrt[3]{|z|} \cos\left(\dfrac{\arg(z)}{3}\right)$

The imaginary parts are cancelled because $u$ and $v$ are complex conjugates.

$y=16 \cos \left(\dfrac{1}{3} \arccos\left(\dfrac{215}{512}\right)+\dfrac{2 \pi k}{3}\right)$ for $k={0,1,2}$ .

Finally,

$x_{1}=\dfrac{16 \cos \left(\dfrac{1}{3} \arccos\left(\dfrac{215}{512}\right)\right)+11}{3} \approx 8.6212$

$x_{2}=\dfrac{16 \cos \left(\dfrac{1}{3} \arccos\left(\dfrac{215}{512}\right)+\dfrac{2 \pi}{3}\right)+11}{3} \approx -0.5201$

$x_{3}=\dfrac{16 \cos \left(\dfrac{1}{3} \arccos\left(\dfrac{215}{512}\right)+\dfrac{4 \pi}{3}\right)+11}{3} \approx 2.8989$

$\lfloor x_1 \rfloor+\lfloor x_2 \rfloor+\lfloor x_3 \rfloor=8-1+2=\boxed{9}$

Try looking for the values of f(x) for a few integers.

For negative integers, the polynomial tends to become 'more' negative as we can easily see. Hence, there would be no negative root. But, f(0) is positive while f(-1) is negative, as a result there has to be a root in between x = 0 and x = -1.

Now, we can look that

$f(1) = 22, f(2) = 15, f(3) = -2$

Hence, there has to be a root between x = 2 and x = 3

Now, for getting the next root, use Vieta,

${a}_{1} + {a}_{2} + {a}_{3} = 11$

${a}_{1}$ would be 2. x (2. x means 2 and '.' means decimal point and x is some number after decimal point)

Similarly, ${a}_{2}$ would be -0. y

Therefore, $2. x - 0. y + {a}_{3} = 11$

Now, as f(3) = -2 and f(2) = 15. it is almost certain that ${a}_{1} > 2.5$ .

As a result, $2. x - 0. y \approx 2. z$

${a}_{3} = 11 - 2. z$

As a result ${a}_{3}$ will be $8. w$

Therefore, $[{a}_{1}] + [{a}_{2}] + [{a}_{3}] = [2. x] + [-0. y] + [8. w] = 2 - 1 + 8 = \boxed{9}$

On taking out the roots of the equation we get three roots

-0.5

2.8

8.6

GIF of -0.5=-1

Of 2.8 =2

Of 8.6=8

Thus we get sum as 8+2+(-1)=9

I used the Newton Raphson algorithm with an initial value of 2 and obtained 2.898927 as one of the roots. Then we use the sum and product of the roots to locate the other 2 roots using the quadratic formula. The roots being 0.650, 2.899 & 7.451 giving values of 0, 2 and 7 and hence 9 as the required answer

$a_1-1 < [a_1] \leq a_1$ ...(1)

$a_2-1 < [a_2] \leq a_2$ ...(2)

$a_3-1 < [a_3] \leq a_3$ ...(3)

Also $a_1+a_2+a_3=11$

adding (1),(2) and (3), we get

$8 < Sum \leq 11$

So answer is 9 , 10 or 11.

u can do this without graphs by mere guessing values and when a sign change occurs between n and n+1 , [root]=n .Then just add up

Well i do not know if there is an algebraic method, i used graph.

we can differentiate the function and check that it has a maxima sligtly to the right of 0 and minima slightly to the leftof 7

Also at x=1 y is positive. And at x=2 negative . so one of the roots GIF is 1

next on putting x=-1 we get negative value

so another roots GIF is -1

now i know that minima lies at very close to 7. it is likely that roots lie at (8,9) or (9,10)

we can check again by sign method that it lies between (9,10) So third GIF is 9.

adding we get 9