Difference between revisions of "Rational root theorem"

Revision as of 16:16, 27 August 2021

In algebra, the rational root theorem states that given an integer polynomial $P(x)$ with leading coefficent $a_n$ and constant term $a_0$ , if $P(x)$ has a rational root $r = \frac pq$ in lowest terms, then $p|a_0$ and $q|a_n$ .

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 $\frac{p}{q}$ be a rational root of $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ , where all $a_r$ are integers; we wish to show that $p|a_0$ and $q|a_n$ . Since $\frac{p}{q}$ is a root of $P(x)$ , $0 = a_n \left(\frac{p}{q}\right)^n + a_{n-1} \left(\frac{p}{q}\right)^{n-1} + \cdots + a_1 \left(\frac{p}{q}\right) + a_0.$ Multiplying by $q^n$ yields $0 = a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p * q^{n-1} + a_0 q^n.$ Using modular arithmetic modulo $p$ , we have $a_0 q^n \equiv 0\pmod p$ , which implies that $p | a_0 q^n$ . Because we've defined $p$ and $q$ to be relatively prime, $\gcd(q^n, p) = 1$ , which implies $p | a_0$ by Euclid's lemma. Via similar logic in modulo $q$ , $q|a_n$ , as required. $\square$

Problems

Here are some problems that are cracked by the rational root theorem.

Problem 1

Factor the polynomial $x^3-5x^2+2x+8$ .

Solution: After testing the divisors of 8, we find that it has roots $-1$ , $2$ , and $4$ . Then because it has leading coefficient $1$ , its factorization is $(x-4)(x-2)(x+1)$ . $\square$

Problem 2

Find all rational roots of the polynomial $x^4-x^3-x^2+x+57$ .

Solution: The polynomial has leading coefficient $1$ and constant term $3 \cdot 19$ , so the rational root theorem guarantees that the only possible rational roots are $-57$ , $-19$ , $-3$ , $-1$ , $1$ , $3$ , $19$ , and $57$ . After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots. $\square$

Problem 3

Using the rational root theorem, prove that $\sqrt{2}$ is irrational.

Solution: The polynomial $x^2 - 2$ has roots $-\sqrt{2}$ and $\sqrt{2}$ . The rational root theorem garuntees that the only possible rational roots are $-2, -1, 1$ , and $2$ . Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because $\sqrt{2}$ is a root of the polynomial, it cannot be a rational number. $\square$

@@ Line 1: / Line 1: @@
-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 lowest terms <math>r = \frac pq</math>, then <math>p|a_0</math> and <math>q|a_n</math>.
+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.
@@ Line 7: / Line 7: @@
 == Problems ==
-Here are some problems that are cracked by the rational root theorem. The answers can be found [[Rational root theorem/problems | here]].
+Here are some problems that are cracked by the rational root theorem.
 === Problem 1 ===

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