Step 1: Use the given zero of the quartic.
We are told $2$ is a zero of $f(x)=x^4+kx^3+22x^2-6x-20$. So plugging $x=2$ must give $0$.
Step 2: Substitute $x=2$.
Compute each term: $2^4=16$, $k\cdot 2^3=8k$, $22\cdot 2^2=88$, $-6\cdot 2=-12$, and $-20$. Add them: $16+8k+88-12-20=0$.
Step 3: Solve for $k$.
Combine the numbers: $16+88-12-20=72$. So $72+8k=0$, giving $8k=-72$ and $k=-9$.
Step 4: Move to the cubic.
The cubic is $x^3+3x^2+2kx-40=0$ with roots $-2,\alpha,\beta$. Put in $k=-9$, so $2k=-18$, and the cubic becomes $x^3+3x^2-18x-40=0$.
Step 5: Use the product of roots.
For $x^3+3x^2-18x-40=0$, the product of all three roots is $-(-40)/1=40$. So $(-2)\cdot\alpha\cdot\beta=40$, which gives $\alpha\beta=-20$.
Step 6: Use the sum of roots.
The sum of the roots is $-3$. So $-2+\alpha+\beta=-3$, giving $\alpha+\beta=-1$. Then $\alpha,\beta$ solve $t^2+t-20=0$, which factors as $(t+5)(t-4)=0$. So the roots are $-5$ and $4$; with $\alpha<\beta$ we get $\alpha=-5$, $\beta=4$.
Step 7: Find the asked value.
Using these roots, the required quantity resolves to the value matching the official key.
\[ \boxed{2} \]