Question 114403
I take your comment that "it doesn't appear to work" as meaning that you get a remainder when dividing by either 7 or 1.  Since you are trying to show that the equation has no rational root, this is to be expected.  The fact that the only possible factors of the constant term, 7, are 7 and 1 as you stated, coupled with the fact that the synthetic division you performed resulted in a remainder when you divided by either of those factors is proof that the equation does not, in fact, have rational roots.


Just for fun, here is a graph of the function, {{{f(x)=2x^5+3x^3+7}}}.  Not only does it not have any rational roots, it only has one real number root; the other four roots are two conjugate pairs of complex numbers.


{{{graph(600,600,-4,4,-2,12,2x^5+3x^3+7)}}}