> .........
> Again, we can evaluate by substituting for the 'x':
>
>    subst q0 1;  // 0.0
>    subst q0 5;  // 36.0

The last two lines should read:

   subst (roots q0) 1;  // 0.0
   subst (roots q0) 5;  // 36.0

The results are the same, but the idea is to use the root-based
representation of the polynomial to evaluate it and thus demonstrate
that (roots q0) is a representation of q0, as it should be.
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