Polynomial: Difference between revisions

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{{#content:Q2150}}
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Then $P(x)$ is a '''polynomial''' over $\mathbb{F}$
then $P(x)$ is a '''polynomial''' of '''degree''' $n$ over $\mathbb{F}$. $a_n$ is said to be its '''leading coefficient''', and $a_nx^n$ its '''leading term'''.


<blockquote>
<blockquote>
Technically, $0$ is a polynomial.
$P(x)=0$ is known as the '''zero polynomial'''.
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</blockquote>


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$x+x^{\frac{1}{2}}$ is '''not''' a polynomial as $\frac{1}{2}\notin \mathbb{N}$
$x+x^{\frac{1}{2}}$ is '''not''' a polynomial as $\frac{1}{2}\notin \mathbb{N}$
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</blockquote>
<ref>{{#cite:Q2204}}</ref>

Revision as of 13:27, 16 October 2026

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Mathematical definition

Let $P(x)$ be a mathematical function $P:\mathbb{F} \to \mathbb{F}$

If $P(x)$, through algebraic manipulation, can be written in the form of

$$P(x)=\sum_{k=0}^{n} a_k x^k, n\in\mathbb{N},a_0, a_1, \dots, a_n\in \mathbb{F}$$

where $x\in \mathbb{F}$ is an independent variable

then $P(x)$ is a polynomial of degree $n$ over $\mathbb{F}$. $a_n$ is said to be its leading coefficient, and $a_nx^n$ its leading term.

$P(x)=0$ is known as the zero polynomial.

$x+x^{\frac{1}{2}}$ is not a polynomial as $\frac{1}{2}\notin \mathbb{N}$

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