Ben is back! 11

Since a root of $f(x)$ is $\sqrt{11}+\sqrt{10}$ , to have integer coefficient, we can assume another positive root is $\sqrt{11}-\sqrt{10}$ . Let the remaining two real roots be $a$ and $b$ , Then, we have:

$\begin{aligned} f(x) & = \left(x-\sqrt{11}-\sqrt{10}\right)\left(x-\sqrt{11}+\sqrt{10}\right)(x-a)(x-b) \\ & = \left((x-\sqrt{11})^2-10\right)(x-a)(x-b) \\ & = \left(x^2 - 2\sqrt{11}x + 1\right) (x-a)(x-b) \end{aligned}$

To have integer coefficients, $(x-a)(x-b) = \left(x^2 + 2\sqrt{11}x + 1\right) = \left(x+\sqrt{11}+\sqrt{10}\right)\left(x+\sqrt{11}-\sqrt{10}\right)$ . Then we have:

$\begin{aligned} f(x) & = \left(x^2 - 2\sqrt{11}x + 1\right) \left(x^2 + 2\sqrt{11}x + 1\right) \\ \implies f(x) & = x^4 - 42x^2+1 & \small \color{#3D99F6} \text{with all integer coefficients} \\ f(1) & = \boxed{-40} \end{aligned}$