jmc

If all the $a_i$ are equal to 0, then the polynomial becomes $x^9=0,$ which has only one integer root, namely $x=0.$ Thus, we can assume that there is some coefficient $a_i$ that is non-zero. Let $k$ be the smallest integer such that $a_k\ne 0$; then we can take out a factor of $x^k,$ to get $$x^k(x^{9-k}+a_8{x}^{8-k}+a_7{x}^{7-k}+\cdots +a_{k+1}x+a_k)=0.$$By the Integer Root Theorem, any integer root of $x^{9-k}+a_8{x}^{8-k}+\cdots +a_{k+1}x+a_k=0$ must divide $a_k=1,$ so the only possible integer roots are 1 and $-1.$ However, if we plug in $x=1,$ we see that $x^{9-k}=1,$ and all the other terms are nonnegative, so $x=1$ cannot be a root.

Therefore, for the original polynomial to have two different integer roots, they must be 0 and $-1.$ For 0 to be a root, it suffices to take $a_0=0,$ and the polynomial is $$x^9+a_8{x}^8+a_7{x}^7+a_6{x}^6+a_5{x}^5+a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x=0.$$We also want $x=-1$ to be a root. We have that $(-1{)}^9=-1,$ so in order for the polynomial to become 0 at $x=-1,$ we must choose some of the $a_i$ to be equal to 1. Specifically, if $k$ is the number of $i$ such that $a_i=1$ and $i$ is odd, then the number of $i$ such that $a_i=1$ and $i$ is even must be $k+1.$

There are four indices that are odd (1, 3, 5, 7), and four indices that are even (2, 4, 6, 8), so the possible values of $k$ are 0, 1, 2, and 3. Furthermore, for each $k,$ so the number of ways to choose $k$ odd indices and $k+1$ even indices is $\left(\genfrac{}{}{0ex}{}{4}{k}\right)\left(\genfrac{}{}{0ex}{}{4}{k+1}\right).$ Therefore, the number of such polynomials is $$\left(\genfrac{}{}{0ex}{}{4}{0}\right)\left(\genfrac{}{}{0ex}{}{4}{1}\right)+\left(\genfrac{}{}{0ex}{}{4}{1}\right)\left(\genfrac{}{}{0ex}{}{4}{2}\right)+\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{4}{3}\right)\left(\genfrac{}{}{0ex}{}{4}{4}\right)=\boxed{56}.$$