Difference between revisions of "Rational Root Theorem"
Williamhu888 (talk | contribs) m (→Proof) |
Anthonyjang (talk | contribs) m (→Problems) |
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===Easy=== | ===Easy=== | ||
− | Factor the polynomial <math>x^3-5x^2+2x+8</math>. | + | 1. Factor the polynomial <math>x^3-5x^2+2x+8</math>. |
===Intermediate=== | ===Intermediate=== | ||
− | Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>. | + | 2. Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>. |
− | Prove that <math>\sqrt{2}</math> is irrational, using the Rational Root Theorem. | + | 3. Prove that <math>\sqrt{2}</math> is irrational, using the Rational Root Theorem. |
+ | |||
+ | ===Answers=== | ||
+ | 1. <math>(x-4)(x-2)(x+1)</math> |
Revision as of 23:28, 7 December 2012
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
, and we are done.
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.