Difference between revisions of "Rational Root Theorem"
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1. <math>(x-4)(x-2)(x+1)</math> | 1. <math>(x-4)(x-2)(x+1)</math> | ||
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2. <math>\text{There are no rational roots for the polynomial.} </math> | 2. <math>\text{There are no rational roots for the polynomial.} </math> | ||
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+ | 3. A polynomial with a root as <math>\sqrt{2}</math> is <math>x^2-2=0</math>. We see that the only possible rational roots are <math>\pm 1</math> and <math>\pm 2</math>, and when substituted, none of these roots work. |
Revision as of 00:40, 17 April 2018
This article is a stub. Help us out by expanding it.
Given a polynomial with integral coefficients,
. The Rational Root Theorem states that if
has a rational root
with
relatively prime positive integers,
is a divisor of
and
is a divisor of
.
As a consequence, every rational root of a monic polynomial with integral coefficients must be integral.
This gives us a relatively quick process to find all "nice" roots of a given polynomial, since given the coefficients we have only a finite number of rational numbers to check.
Contents
[hide]Proof
Given is a rational root of a polynomial
, where the
's are integers, we wish to show that
and
. Since
is a root,
Multiplying by
, we have:
Examining this in modulo
, we have
. As
and
are relatively prime,
. With the same logic, but with modulo
, we have
, which completes the proof.
Problems
Easy
1. Factor the polynomial .
Intermediate
2. Find all rational roots of the polynomial .
3. Prove that is irrational, using the Rational Root Theorem.
Answers
1.
2.
3. A polynomial with a root as is
. We see that the only possible rational roots are
and
, and when substituted, none of these roots work.