Administrator

Hello!
Great question!
So let me think out loud for a minute.
So the idea is that when you divide by x=c, your polynomial is:
p(x) = (xc)(q(x)) + r
Let's say the signs of the division tableau are all () and I pick k > c > 0.
Then p(k) = (kc)(q(k)) + r
Now (kc) > 0 so (kc) is (+)
q(k) is a sum of terms which are powers of k (all (+)) times negative coefficients. So that means q(k) is ().
Finally, r is () as well.
So p(k) = (+)() + () = () which means k can't be a zero.
Let know what you think!
Carl
