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 coefficent <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root in lowest terms <math>r = \frac pq</math>, <math>p|a_0</math> and <math>q|a_n</math>. | + | In [[algebra]], the '''rational root theorem''' states that given an integer [[polynomial]] <math>P(x)</math> with leading coefficent <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root in lowest terms <math>r = \frac pq</math>, then <math>p|a_0</math> and <math>q|a_n</math>. |
This theorem aids significantly at finding the "nice" roots of a given polynomial, since the coefficients entail only a finite amount of rational numbers to check as roots. | This theorem aids significantly at finding the "nice" roots of a given polynomial, since the coefficients entail only a finite amount of rational numbers to check as roots. | ||
== Proof == | == Proof == | ||
− | Let <math>\frac{p}{q}</math> be a rational root of <math>P(x)= | + | 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 all <math>a_r</math> are integers; 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 == | == Problems == |
Revision as of 13:05, 15 July 2021
In algebra, the rational root theorem states that given an integer polynomial with leading coefficent
and constant term
, if
has a rational root in lowest terms
, then
and
.
This theorem aids significantly at finding the "nice" roots of a given polynomial, since the coefficients entail only a finite amount of rational numbers to check as roots.
Proof
Let be a rational root of
, where all
are integers; 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.
Problems
Here are some problems that are cracked by the rational root theorem. The answers can be found here.
Introductory
- Factor the polynomial
.
Intermediate
- Find all rational roots of the polynomial
.
- Prove that
is irrational, using the Rational Root Theorem.
Answers
1.
2.
3. A polynomial with integer coefficients and has a root as
must also have
as a root. The simplest polynomial is
which is
. We see that the only possible rational roots are
and
, and when substituted, none of these roots work.