Difference between revisions of "Rational root theorem"
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=== Example 3 === | === Example 3 === | ||
− | ''Using the rational root | + | ''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> | '''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> |
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.