Difference between revisions of "Rational Root Theorem"
I like pie (talk | contribs) m |
(→Intermediate) |
||
(18 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
− | + | ||
Line 7: | Line 7: | ||
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. | 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. | ||
+ | |||
+ | == Proof == | ||
+ | |||
+ | Given <math>\frac{p}{q}</math> is a rational root of a polynomial <math>f(x)=a_nx^n+x_{n-1}x^{n-1}+\cdots +a_0</math>, where the <math>a_n</math>'s are integers, 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, <cmath>0=a_n\left(\frac{p}{q}\right)^n+\cdots +a_0</cmath> Multiplying by <math>q^n</math>, we have: <cmath>0=a_np^n+a_{n-1}p^{n-1}q+\cdots+a_0q^n</cmath> Examining this in modulo <math>p</math>, we have <math>a_0q^n\equiv 0\pmod p</math>. As <math>q</math> and <math>p</math> are relatively prime, <math>p|a_0</math>. With the same logic, but with modulo <math>q</math>, we have <math>q|a_n</math>, which completes the proof. | ||
==Problems== | ==Problems== | ||
+ | |||
+ | ===Easy=== | ||
+ | |||
+ | 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>. |
+ | |||
+ | 3. Prove that <math>\sqrt{2}</math> is irrational, using the Rational Root Theorem. | ||
+ | |||
+ | ===Answers=== | ||
+ | 1. <math>(x-4)(x-2)(x+1)</math> | ||
+ | |||
+ | 2. <math>\text{There are no rational roots for the polynomial.} </math> | ||
+ | |||
+ | 3. A polynomial with integer coefficients and has a root as <math>\sqrt{2}</math> must also have <math>-\sqrt{2}</math> as a root. The simplest polynomial is <math>(x+\sqrt{2})(x-\sqrt{2})</math> which 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. | ||
+ | {{stub}} |
Revision as of 13:46, 25 September 2020
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 , 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 integer coefficients and has a root as must also have as a root. The simplest polynomial is which is . We see that the only possible rational roots are and , and when substituted, none of these roots work. This article is a stub. Help us out by expanding it.