jmc

Let's deal with the restriction first.

The restriction is that we have to place a math book on both ends. We have 3 choices for the math book to place on the left end, and then 2 choices for the math book to place on the right end.

Then we simply need to arrange the other 6 books in the middle. This is a basic permutation problem, so there are $6!$ ways to arrange the 6 remaining books.

So there are a total of $3\times 2\times 6!=\boxed{4,320}$ ways to arrange the books on the bookshelf.