jmc

The roots of the corresponding equation $x^2+bx+2=0$ are $$\frac{-b\pm \sqrt{b^2-8}}{2}.$$(Note that these roots must be real, otherwise, the inequality $x^2+bx+2\le 0$ has no real solutions.) Thus, the solution to this inequality $x^2+bx+2\le 0$ is $$\frac{-b-\sqrt{b^2-8}}{2}\le x\le \frac{-b+\sqrt{b^2-8}}{2}.$$If the length of this interval is at least 4, then it must contain at least 4 integers, so the width of this interval must be less than 4. Thus, $$\sqrt{b^2-8}<4.$$Then $b^2-8<16,$ so $b^2<24.$ We must also have $b^2>8.$ The only possible values of $b$ are then $-4,$ $-3,$ 3, and 4. We can look at each case.

$$\begin{array}{cc}b& \text{Integer solutions to x2+bx+2≤0}\\ -4& 1,2,3\\ -3& 1,2\\ 3& -2,-1\\ 4& -3,-2,-1\end{array}$$Thus, there are $\boxed{2}$ values of $b$ that work, namely $-4$ and 4.