Difference between revisions of "Rational root theorem"
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− | In [[algebra]], the '''rational root theorem''' states that given an integer [[polynomial]] <math>P(x)</math> with leading coefficent <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root | + | In [[algebra]], the '''rational root theorem''' states that given an integer [[polynomial]] <math>P(x)</math> with leading coefficent <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root <math>r = \frac pq</math> in lowest terms, then <math>p|a_0</math> and <math>q|a_n</math>. |
This theorem aids significantly at finding the "nice" roots of a given polynomial, since the coefficients entail only a finite amount of rational numbers to check as roots. | This theorem aids significantly at finding the "nice" roots of a given polynomial, since the coefficients entail only a finite amount of rational numbers to check as roots. | ||
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== Problems == | == Problems == | ||
− | Here are some problems that are cracked by the rational root theorem | + | Here are some problems that are cracked by the rational root theorem. |
=== Problem 1 === | === Problem 1 === |
Revision as of 15:16, 27 August 2021
In algebra, the rational root theorem states that given an integer polynomial with leading coefficent and constant term , if has a rational root in lowest terms, then and .
This theorem aids significantly at finding the "nice" roots of a given polynomial, since the coefficients entail only a finite amount of rational numbers to check as roots.
Proof
Let be a rational root of , where all are integers; 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.
Problems
Here are some problems that are cracked by the rational root theorem.
Problem 1
Factor the polynomial .
Solution: After testing the divisors of 8, we find that it has roots , , and . Then because it has leading coefficient , its factorization is .
Problem 2
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.
Problem 3
Using the rational root theorem, prove that is irrational.
Solution: The polynomial has roots and . The rational root theorem garuntees that the only possible rational roots 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.