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 | + | 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>. |
− | This theorem | + | This theorem is most often used to guess the roots of polynomials. It sees widespread usage in introductory and intermediate mathematics competitions. |
== 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 | + | 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 | |
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− | === | + | == Examples == |
− | + | Here are some problems with solutions that utilize the rational root theorem. | |
− | |||
− | === | + | === Example 1 === |
− | + | ''Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>. | |
− | + | ||
− | 3. | + | '''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> |
+ | |||
+ | === Example 2 === | ||
+ | ''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), x={-1, 2, 4}</math>. <math>\square</math> | ||
+ | |||
+ | === Example 3 === | ||
+ | ''Using the rational root theorem, prove that <math>\sqrt{2}</math> is irrational.'' | ||
+ | |||
+ | '''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 13:07, 20 February 2024
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: 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 .
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.