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Difference between revisions of "Factor Theorem"
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In algebra, the '''Factor theorem''' is a theorem regarding the relationships between the factors of a polynomial and its roots.
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''This page is under heavy construction--[[User:10000th User|10000th User]] 13:44, 15 November 2007 (EST)''
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== Introduction ==
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One of it's most important applications is if you are given that a polynomial have certain roots, you will know certain linear factors of the polynomial. Thus, you can test if a linear factor is a factor of a polynomial without using polynomial division and instead plugging in numbers. Conversely, you can determine whether a number in the form <math>f(a)</math> (<math>a</math> is constant, <math>f</math> is polynomial) is <math>0</math> using polynomial division rather than plugging in large values.
  
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== Theorem and Proof ==
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==Statement==
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'''Theorem:''' If <math>P(x)</math> is a polynomial, then <math>x-a</math> is a factor <math>P(x)</math> iff <math>P(a)=0</math>.
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The '''Factor Theorem''' says that if <math>P(x)</math> is a [[polynomial]], then <math>x-a</math> is a [[factor]] of <math>P(x)</math> if and only if <math>P(a)=0</math>.
  
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*'''Proof:''' If <math>x - a</math> is a factor of <math>P(x)</math>, then <math>P(x) = (x - a)Q(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math>. Then <math>P(a) = (a - a)Q(a) = 0</math>.
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==Proof==
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If <math>x - a</math> is a factor of <math>P(x)</math>, then <math>P(x) = (x - a)Q(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math>. Then <math>P(a) = (a - a)Q(a) = 0</math>.
  
 
Now suppose that <math>P(a) = 0</math>.
 
Now suppose that <math>P(a) = 0</math>.
  
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Apply division algorithm to get <math>P(x) = (x - a)Q(x) + R(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math> and <math>R(x)</math> is the remainder polynomial such that <math>0\le\deg(R(x)) < \deg(x - a) = 1</math>.
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Apply [[Remainder Theorem]] to get <math>P(x) = (x - a)Q(x) + R(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math> and <math>R(x)</math> is the remainder polynomial such that <math>0\le\deg(R(x)) < \deg(x - a) = 1</math>. This means that <math>R(x)</math> can be at most a [[constant]] polynomial.
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Substitute <math>x = a</math> and get <math>P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0</math>. Since <math>R(x)</math> is a  constant polynomial, <math>R(x) = 0</math> for all <math>x</math>.
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Therefore, <math>P(x) = (x - a)Q(x)</math>, which shows that <math>x - a</math> is a factor of <math>P(x)</math>.
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==Problems==
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Here are some problems that can be solved using the Factor Theorem:
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===Introductory===
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===Intermediate===
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Suppose <math>f(x)</math> is a <math>10000000010</math>-degree polynomial. The Fundamental Theorem of Algebra tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \cdots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that
  
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This means that <math>R(x)</math> can be at most a constant polynomial.
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<math>(2+r_1)(2+r_2) \cdots (2+r_{10000000010})=m!</math>
  
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Substitute <math>x = a</math> and get <math>P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0</math>.
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for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of
  
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But <math>R(x)</math> is a constant polynomial and so <math>R(x) = 0</math> for all <math>x</math>.
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<math>(1+r_1)(1+r_2) \cdots (1+r_{10000000010})</math>.
  
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Therefore, <math>P(x) = (x - a)Q(x)</math>, which shows that <math>x - a</math> is a factor of <math>P(x)</math>.
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Find the number of factors of the prime <math>999999937</math> in <math>p</math>. (Source: I made it. Solution [[User:Ddk001#Problem_7|here]])
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===Olympaid===
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If <math>P(x)</math> denotes a polynomial of degree <math>n</math> such that<cmath>P(k)=\frac{k}{k+1}</cmath>for <math>k=0,1,2,\cdots,n</math>, determine <math>P(n+1)</math>.
  
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(Source: [[1975 USAMO Problems/Problem 3|1975 USAMO Problem 3]])
  
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== Problems ==
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==See Also==
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*[[Polynomials]]
  
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== See also ==
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*[[Remainder Theorem]]
  
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[[Category: Polynomials]]
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[[Category:Algebra]]
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[[Category:Polynomials]]
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[[Category:Theorems]]
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[[Category:Mathematics]]
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Latest revision as of 17:04, 28 September 2024

In algebra, the Factor theorem is a theorem regarding the relationships between the factors of a polynomial and its roots.


One of it's most important applications is if you are given that a polynomial have certain roots, you will know certain linear factors of the polynomial. Thus, you can test if a linear factor is a factor of a polynomial without using polynomial division and instead plugging in numbers. Conversely, you can determine whether a number in the form $f(a)$ ($a$ is constant, $f$ is polynomial) is $0$ using polynomial division rather than plugging in large values.

Statement

The Factor Theorem says that if $P(x)$ is a polynomial, then $x-a$ is a factor of $P(x)$ if and only if $P(a)=0$.

Proof

If $x - a$ is a factor of $P(x)$, then $P(x) = (x - a)Q(x)$, where $Q(x)$ is a polynomial with $\deg(Q(x)) = \deg(P(x)) - 1$. Then $P(a) = (a - a)Q(a) = 0$.

Now suppose that $P(a) = 0$.

Apply Remainder Theorem to get $P(x) = (x - a)Q(x) + R(x)$, where $Q(x)$ is a polynomial with $\deg(Q(x)) = \deg(P(x)) - 1$ and $R(x)$ is the remainder polynomial such that $0\le\deg(R(x)) < \deg(x - a) = 1$. This means that $R(x)$ can be at most a constant polynomial.

Substitute $x = a$ and get $P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0$. Since $R(x)$ is a constant polynomial, $R(x) = 0$ for all $x$.

Therefore, $P(x) = (x - a)Q(x)$, which shows that $x - a$ is a factor of $P(x)$.

Problems

Here are some problems that can be solved using the Factor Theorem:

Introductory

Intermediate

Suppose $f(x)$ is a $10000000010$-degree polynomial. The Fundamental Theorem of Algebra tells us that there are $10000000010$ roots, say $r_1, r_2, \cdots, r_{10000000010}$. Suppose all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$. Also, suppose that

$(2+r_1)(2+r_2) \cdots (2+r_{10000000010})=m!$

for an integer $m$. If $p$ is the minimum possible positive integral value of

$(1+r_1)(1+r_2) \cdots (1+r_{10000000010})$.

Find the number of factors of the prime $999999937$ in $p$. (Source: I made it. Solution here)

Olympaid

If $P(x)$ denotes a polynomial of degree $n$ such that\[P(k)=\frac{k}{k+1}\]for $k=0,1,2,\cdots,n$, determine $P(n+1)$.

(Source: 1975 USAMO Problem 3)

See Also

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