Difference between revisions of "Rational Root Theorem"
Williamhu888 (talk | contribs) m (→Proof) |
Anthonyjang (talk | contribs) m (→Problems) |
||
Line 16: | Line 16: | ||
===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 22:28, 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
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.