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Difference between revisions of "Rational root theorem"
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== 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>
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Intro to Rational Roots theorem: https://www.youtube.com/shorts/wKpmfnyKeeM
  
 
== Examples ==
 
== Examples ==
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Here are some problems that are cracked by the rational root theorem.
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Here are some problems with solutions that utilize the rational root theorem.
  
 
=== Example 1 ===
 
=== Example 1 ===
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''Factor the polynomial <math>x^3-5x^2+2x+8</math>.''
 
''Factor the polynomial <math>x^3-5x^2+2x+8</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)</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>
  
 
=== Example 3 ===
 
=== 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.''
  
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'''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

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