Difference between revisions of "Rational root theorem"

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In [[algebra]], the '''rational root theorem''' states that given an integer [[polynomial]] <math>P(x)</math> with leading coefficient <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root <math>r = \frac pq</math> in lowest terms, then <math>p|a_0</math> and <math>q|a_n</math>.
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In [[algebra]], the '''rational root theorem''' states that given an integer [[polynomial]] <math>P(x)</math> with leading coefficient <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root <math>r = p/q</math> in lowest terms, then <math>p|a_0</math> and <math>q|a_n</math>.
  
Using this theorem, it is often possible to guess the roots of a polynomial with a degree of accuracy.
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This theorem is most often used to guess the roots of polynomials. It sees widespread usage in introductory and intermediate mathematics competitions.
 
 
This theorem helps to find the "nice" roots of a polynomial significantly, aiding to how it limits the possible rational roots to a finite number.
 
  
 
== Proof ==
 
== Proof ==
 
Let <math>\frac{p}{q}</math> be a rational root of <math>P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0</math>, where every <math>a_r</math> is an integer; we wish to show that <math>p|a_0</math> and <math>q|a_n</math>. Since <math>\frac{p}{q}</math> is a root of <math>P(x)</math>, <cmath>0 = a_n \left(\frac{p}{q}\right)^n + a_{n-1} \left(\frac{p}{q}\right)^{n-1} + \cdots + a_1 \left(\frac{p}{q}\right) + a_0.</cmath> Multiplying by <math>q^n</math> yields <cmath>0 = a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p * q^{n-1} + a_0 q^n.</cmath> Using [[modular arithmetic]] modulo <math>p</math>, we have <math>a_0 q^n \equiv 0\pmod p</math>, which implies that <math>p | a_0 q^n</math>. Because we've defined <math>p</math> and <math>q</math> to be relatively prime, <math>\gcd(q^n, p) = 1</math>, which implies <math>p | a_0</math> by [[Euclid's lemma]]. Via similar logic in modulo <math>q</math>, <math>q|a_n</math>, as required. <math>\square</math>
 
Let <math>\frac{p}{q}</math> be a rational root of <math>P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0</math>, where every <math>a_r</math> is an integer; we wish to show that <math>p|a_0</math> and <math>q|a_n</math>. Since <math>\frac{p}{q}</math> is a root of <math>P(x)</math>, <cmath>0 = a_n \left(\frac{p}{q}\right)^n + a_{n-1} \left(\frac{p}{q}\right)^{n-1} + \cdots + a_1 \left(\frac{p}{q}\right) + a_0.</cmath> Multiplying by <math>q^n</math> yields <cmath>0 = a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p * q^{n-1} + a_0 q^n.</cmath> Using [[modular arithmetic]] modulo <math>p</math>, we have <math>a_0 q^n \equiv 0\pmod p</math>, which implies that <math>p | a_0 q^n</math>. Because we've defined <math>p</math> and <math>q</math> to be relatively prime, <math>\gcd(q^n, p) = 1</math>, which implies <math>p | a_0</math> by [[Euclid's lemma]]. Via similar logic in modulo <math>q</math>, <math>q|a_n</math>, as required. <math>\square</math>
  
== Problems ==
 
Here are some problems that are cracked by the rational root theorem.
 
  
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Intro to Rational Roots theorem: https://www.youtube.com/shorts/wKpmfnyKeeM
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== Examples ==
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Here are some problems with solutions that utilize the rational root theorem.
  
=== Problem 1 ===
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=== Example 1 ===
 
''Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>.
 
''Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>.
  
 
'''Solution''': The polynomial has leading coefficient <math>1</math> and constant term <math>3 \cdot 19</math>, so the rational root theorem guarantees that the only possible rational roots are <math>-57</math>, <math>-19</math>, <math>-3</math>, <math>-1</math>, <math>1</math>, <math>3</math>, <math>19</math>, and <math>57</math>. After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots. <math>\square</math>
 
'''Solution''': The polynomial has leading coefficient <math>1</math> and constant term <math>3 \cdot 19</math>, so the rational root theorem guarantees that the only possible rational roots are <math>-57</math>, <math>-19</math>, <math>-3</math>, <math>-1</math>, <math>1</math>, <math>3</math>, <math>19</math>, and <math>57</math>. After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots. <math>\square</math>
  
=== Problem 2 ===
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=== Example 2 ===
 
''Factor the polynomial <math>x^3-5x^2+2x+8</math>.''
 
''Factor the polynomial <math>x^3-5x^2+2x+8</math>.''
  
'''Solution''': After testing the divisors of 8, we find that it has roots <math>-1</math>, <math>2</math>, and <math>4</math>. Then because it has leading coefficient <math>1</math>, the [[factor theorem]] tells us that it has the factorization <math>(x-4)(x-2)(x+1)</math>. <math>\square</math>
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'''Solution''': After testing the divisors of 8, we find that it has roots <math>-1</math>, <math>2</math>, and <math>4</math>. Then because it has leading coefficient <math>1</math>, the [[factor theorem]] tells us that it has the factorization <math>(x-4)(x-2)(x+1), x={-1, 2, 4}</math>. <math>\square</math>
  
=== Problem 3 ===
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=== Example 3 ===
 
''Using the rational root theorem, prove that <math>\sqrt{2}</math> is irrational.''
 
''Using the rational root theorem, prove that <math>\sqrt{2}</math> is irrational.''
  
'''Solution''': The polynomial <math>x^2 - 2</math> has roots <math>-\sqrt{2}</math> and <math>\sqrt{2}</math>. The rational root theorem garuntees that the only possible rational roots of this polynomial are <math>-2, -1, 1</math>, and <math>2</math>. Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because <math>\sqrt{2}</math> is a root of the polynomial, it cannot be a rational number. <math>\square</math>
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'''Solution''': The polynomial <math>x^2 - 2</math> has roots <math>\pm \sqrt{2}</math>. The rational root theorem guarantees that the only possible rational roots of this polynomial are <math>-2, -1, 1</math>, and <math>2</math>. Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because <math>\sqrt{2}</math> is a root of the polynomial, it cannot be a rational number. <math>\square</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 12:07, 20 February 2024

In algebra, the rational root theorem states that given an integer polynomial $P(x)$ with leading coefficient $a_n$ and constant term $a_0$, if $P(x)$ has a rational root $r = p/q$ in lowest terms, then $p|a_0$ and $q|a_n$.

This theorem is most often used to guess the roots of polynomials. It sees widespread usage in introductory and intermediate mathematics competitions.

Proof

Let $\frac{p}{q}$ be a rational root of $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$, where every $a_r$ is an integer; we wish to show that $p|a_0$ and $q|a_n$. Since $\frac{p}{q}$ is a root of $P(x)$, \[0 = a_n \left(\frac{p}{q}\right)^n + a_{n-1} \left(\frac{p}{q}\right)^{n-1} + \cdots + a_1 \left(\frac{p}{q}\right) + a_0.\] Multiplying by $q^n$ yields \[0 = a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p * q^{n-1} + a_0 q^n.\] Using modular arithmetic modulo $p$, we have $a_0 q^n \equiv 0\pmod p$, which implies that $p | a_0 q^n$. Because we've defined $p$ and $q$ to be relatively prime, $\gcd(q^n, p) = 1$, which implies $p | a_0$ by Euclid's lemma. Via similar logic in modulo $q$, $q|a_n$, as required. $\square$


Intro to Rational Roots theorem: https://www.youtube.com/shorts/wKpmfnyKeeM

Examples

Here are some problems with solutions that utilize the rational root theorem.

Example 1

Find all rational roots of the polynomial $x^4-x^3-x^2+x+57$.

Solution: The polynomial has leading coefficient $1$ and constant term $3 \cdot 19$, so the rational root theorem guarantees that the only possible rational roots are $-57$, $-19$, $-3$, $-1$, $1$, $3$, $19$, and $57$. After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots. $\square$

Example 2

Factor the polynomial $x^3-5x^2+2x+8$.

Solution: After testing the divisors of 8, we find that it has roots $-1$, $2$, and $4$. Then because it has leading coefficient $1$, the factor theorem tells us that it has the factorization $(x-4)(x-2)(x+1), x={-1, 2, 4}$. $\square$

Example 3

Using the rational root theorem, prove that $\sqrt{2}$ is irrational.

Solution: The polynomial $x^2 - 2$ has roots $\pm \sqrt{2}$. The rational root theorem guarantees that the only possible rational roots of this polynomial are $-2, -1, 1$, and $2$. Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because $\sqrt{2}$ is a root of the polynomial, it cannot be a rational number. $\square$

See also