0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Have you tried induction? $\endgroup$ – Clement C. Mar 31 at 21:28
  • 2
    $\begingroup$ If you already know that $z$ is a root iff $x-z$ divides $p$ then I really don't understand what is the problem. $\endgroup$ – Mark Mar 31 at 21:28
  • $\begingroup$ Don't forget that $\deg(fg)=\deg(f)+\deg(g)$. $\endgroup$ – Bernard Mar 31 at 21:56
  • $\begingroup$ 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 $\endgroup$ – Vasya Mar 31 at 21:57
  • $\begingroup$ @Vasya A.k.a., "induction." $\endgroup$ – Clement C. Mar 31 at 21:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.