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: [url]https://www.youtube.com/shorts/wKpmfnyKeeM[/url] | ||
== Examples == | == Examples == |
Revision as of 13:49, 22 June 2023
In algebra, the rational root theorem states that given an integer polynomial with leading coefficient
and constant term
, if
has a rational root
in lowest terms, then
and
.
This theorem is most often used to guess the roots of polynomials. It sees widespread usage in introductory and intermediate mathematics competitions.
Proof
Let be a rational root of
, where every
is an integer; we wish to show that
and
. Since
is a root of
,
Multiplying by
yields
Using modular arithmetic modulo
, we have
, which implies that
. Because we've defined
and
to be relatively prime,
, which implies
by Euclid's lemma. Via similar logic in modulo
,
, as required.
Intro to Rational Roots theorem: [url]https://www.youtube.com/shorts/wKpmfnyKeeM[/url]
Examples
Here are some problems with solutions that utilize the rational root theorem.
Example 1
Find all rational roots of the polynomial .
Solution: The polynomial has leading coefficient and constant term
, so the rational root theorem guarantees that the only possible rational roots are
,
,
,
,
,
,
, and
. After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots.
Example 2
Factor the polynomial .
Solution: After testing the divisors of 8, we find that it has roots ,
, and
. Then because it has leading coefficient
, the factor theorem tells us that it has the factorization
.
Example 3
Using the rational root theorem, prove that is irrational.
Solution: The polynomial has roots
. The rational root theorem guarantees that the only possible rational roots of this polynomial are
, and
. Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because
is a root of the polynomial, it cannot be a rational number.