# Proof: A (real) polynomial of degree d has at most d (real) roots

A (real) polynomial of degree $$d$$ has at most $$d$$ (real) roots

The catch is the only things I can cite are

1) The Division Algorithm for Polynomials

2) Prop 6.19: Let p(x) be a real polynomial. The number z is a root of p(x) iff there exists polynomial q(x) such that p(x)=(x-z)q(x).

Any help appreciated.

• Have you tried induction? – Clement C. Mar 31 at 21:28
• If you already know that $z$ is a root iff $x-z$ divides $p$ then I really don't understand what is the problem. – Mark Mar 31 at 21:28
• Don't forget that $\deg(fg)=\deg(f)+\deg(g)$. – Bernard Mar 31 at 21:56
• what is the degree of the polynomial $q(x)$? The next step is apply proposition to $q(x)$ and so forth until you reach polynomial of zero degree – Vasya Mar 31 at 21:57
• @Vasya A.k.a., "induction." – Clement C. Mar 31 at 21:58