Polynomial - Revision history

Rongzhou: /* Properties */

2026-10-16T13:44:23Z

Properties

← Older revision		Revision as of 13:44, 16 October 2026
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	<ref>{{#cite:Q2204}}</ref>		<ref>{{#cite:Q2204}}</ref>

			== Properties ==

			=== Composition of polynomials ===

			If $P_1(x)=P_2(x)+P_3(x)$, then $\deg(P_1)=\max[\deg(P_2),\deg(P_3)]$.

			If $P_1(x)=P_2(x)P_3(x)$, then $\deg(P_1)=\deg(P_2)+\deg(P_3)$

			<blockquote>
			To avoid breaking this property, it is defined that $P(x)=0$ has degree $-\infty$.
			</blockquote>

Rongzhou: /* Mathematical definition */

2026-10-16T13:41:46Z

Mathematical definition

← Older revision		Revision as of 13:41, 16 October 2026
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	{{#content:Q2150}}		{{#content:Q2150}}

	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'''.		then $P(x)$ is a '''polynomial''' over $\mathbb{F}$. If $P(x) \neq 0$, it has '''degree''' $n$, and $a_n,a_n x^n$ are said to be its '''leading coefficient''' and '''leading term''' respectively.

	<blockquote>		<blockquote>
	$P(x)=0$ is known as the '''zero polynomial'''.		$P(x)=0$ is known as the '''zero polynomial'''. Its degree is customarily defined as $-\infty$. Here's [[#Composition of polynomials\|why]].
	</blockquote>		</blockquote>

Rongzhou: /* Mathematical definition */

2026-10-16T13:27:11Z

Mathematical definition

← Older revision		Revision as of 13:27, 16 October 2026
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	{{#content:Q2150}}		{{#content:Q2150}}

	~~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'''.
	</blockquote>		</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 $\frac{1}{2}\notin \mathbb{N}$
	</blockquote>		</blockquote>

			<ref>{{#cite:Q2204}}</ref>

Rongzhou: /* Degree, leading coefficient/term */

2026-10-16T13:23:16Z

Degree, leading coefficient/term

← Older revision		Revision as of 13:23, 16 October 2026
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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 $\frac{1}{2}\notin \mathbb{N}$
	</blockquote>		</blockquote>

	~~== Degree, leading coefficient/term ==~~

	~~The degree of a nonzero polynomial $P(x) \, \deg(P)=\max\{k: a_k\ne 0\}$.~~

Rongzhou: /* Degree */

2026-10-16T13:22:22Z

Degree

← Older revision		Revision as of 13:22, 16 October 2026
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	</blockquote>		</blockquote>

	== Degree ==		== Degree, leading coefficient/term ==

	The degree of a nonzero polynomial $P(x) \, \deg P=\max\{k: a_k\ne 0\}$.		The degree of a nonzero polynomial $P(x) \, \deg(P)=\max\{k: a_k\ne 0\}$.

Rongzhou: Created page with "== 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 {{#content:Q2150}} Then $P(x)$ is a '''polynomial''' over $\mathbb{F}$
Technically, $0$ is a polynomial.

$x+x^{\frac{1}{2}}$ is '''not''' a polynomial as $\frac{1}{2}\notin \mathbb{N}$
== Degree == The degree of a nonzero polynomial $P(x) \,..."

2026-10-13T06:29:28Z

Created page with "== 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 {{#content:Q2150}} Then $P(x)$ is a '''polynomial''' over $\mathbb{F}$ <blockquote> Technically, $0$ is a polynomial. </blockquote> <blockquote> $x+x^{\frac{1}{2}}$ is '''not''' a polynomial as $\frac{1}{2}\notin \mathbb{N}$ </blockquote> == Degree == The degree of a nonzero polynomial $P(x) \,..."

New page

== 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

{{#content:Q2150}}

Then $P(x)$ is a '''polynomial''' over $\mathbb{F}$

<blockquote>
Technically, $0$ is a polynomial.
</blockquote>

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

== Degree ==

The degree of a nonzero polynomial $P(x) \, \deg P=\max\{k: a_k\ne 0\}$.