Exercise 5.5

[rswinkle]

-- Sure, if this is what you mean.

function polyval (p, x)
   local sum, pow = 0, 1
   for _, a in ipairs (p) do
      sum = sum + a * pow
      pow = pow * x
   end
   return sum
end

That would still be 2(n+1) multiplications. They mean this:
a0 + x*(a1 + x*(a2 + x*(a3 ...+ x*an)))))

which would look something like this:

function poly_eval2(p, x)
  ret = p[#p]
  for k = #p-1,1,-1 do
    ret = ret * x
    ret = ret + p[k]
  end
  return ret
end