A polynomial of degree n has at most n zeros

Dependencies:

  1. Field
  2. Factor theorem
  3. Product of linear factors is a factor
  4. Degree of factor is less than degree of polynomial

Let $F$ be a field. Let $p(x) \in F[x]$ and $\deg(p) = n$. Then $p$ has at most $n$ zeros.

Proof

Let $a_1, a_2, \ldots, a_k$ be zeros of $p$. By factor theorem, $(x-a_i) \mid p(x)$ for all $1 \le i \le k$. This means $r(x) = \prod_{i=1}^k (x - a_i)$ divides $p(x)$.

Since $r$ is a factor of $p$, $\deg(r) \le \deg(p) \implies k \le n$.

Dependency for: None

Info:

Transitive dependencies:

  1. Group
  2. Ring
  3. Polynomial
  4. Polynomial divisibility
  5. Degree of sum of polynomials
  6. Degree of product of polynomials
  7. Degree of factor is less than degree of polynomial
  8. Zero divisors of a polynomial
  9. Field
  10. 0x = 0 = x0
  11. Integral Domain
  12. A field is an integral domain
  13. Polynomial division theorem
  14. Factor theorem
  15. Product of linear factors is a factor