Yes, what I copied is what the book says. What's weird is I can plug in numbers for the integers to that meet the requirements set and show that -1 can be a root.
The equation was xn+p1xn-1+p2xn-2+...+pn = 0.
Let n=2, x=-1, p1=2, p2=1 (because pn was required to be 1)
p1+p2=2+1\neq-1
It meets...