A polynomial of degree n has at most n zeros
Dependencies:
- Field
- Factor theorem
- Product of linear factors is a factor
- 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:
- Depth: 8
- Number of transitive dependencies: 15
Transitive dependencies:
- Group
- Ring
- Polynomial
- Degree of product of polynomials
- Zero divisors of a polynomial
- Degree of factor is less than degree of polynomial
- Polynomial divisibility
- Degree of sum of polynomials
- Integral Domain
- 0x = 0 = x0
- Field
- A field is an integral domain
- Polynomial division theorem
- Factor theorem
- Product of linear factors is a factor