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 coefficient <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root <math>r = | + | In [[algebra]], the '''rational root theorem''' states that given an integer [[polynomial]] <math>P(x)</math> with leading coefficient <math>a_n</math> and constant term <math>a_0</math>, if <math>P(x)</math> has a rational root <math>r = p/q</math> in lowest terms, then <math>p|a_0</math> and <math>q|a_n</math>. |
− | + | This theorem is most often used to guess the roots of polynomials. It sees widespread usage in introductory and intermediate mathematics competitions. | |
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== Proof == | == Proof == | ||
Let <math>\frac{p}{q}</math> be a rational root of <math>P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0</math>, where every <math>a_r</math> is an integer; 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 of <math>P(x)</math>, <cmath>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.</cmath> Multiplying by <math>q^n</math> yields <cmath>0 = a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p * q^{n-1} + a_0 q^n.</cmath> Using [[modular arithmetic]] modulo <math>p</math>, we have <math>a_0 q^n \equiv 0\pmod p</math>, which implies that <math>p | a_0 q^n</math>. Because we've defined <math>p</math> and <math>q</math> to be relatively prime, <math>\gcd(q^n, p) = 1</math>, which implies <math>p | a_0</math> by [[Euclid's lemma]]. Via similar logic in modulo <math>q</math>, <math>q|a_n</math>, as required. <math>\square</math> | Let <math>\frac{p}{q}</math> be a rational root of <math>P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0</math>, where every <math>a_r</math> is an integer; 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 of <math>P(x)</math>, <cmath>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.</cmath> Multiplying by <math>q^n</math> yields <cmath>0 = a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p * q^{n-1} + a_0 q^n.</cmath> Using [[modular arithmetic]] modulo <math>p</math>, we have <math>a_0 q^n \equiv 0\pmod p</math>, which implies that <math>p | a_0 q^n</math>. Because we've defined <math>p</math> and <math>q</math> to be relatively prime, <math>\gcd(q^n, p) = 1</math>, which implies <math>p | a_0</math> by [[Euclid's lemma]]. Via similar logic in modulo <math>q</math>, <math>q|a_n</math>, as required. <math>\square</math> | ||
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+ | Intro to Rational Roots theorem: https://www.youtube.com/shorts/wKpmfnyKeeM | ||
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+ | == Examples == | ||
+ | Here are some problems with solutions that utilize the rational root theorem. | ||
− | === | + | === Example 1 === |
''Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>. | ''Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>. | ||
'''Solution''': The polynomial has leading coefficient <math>1</math> and constant term <math>3 \cdot 19</math>, so the rational root theorem guarantees that the only possible rational roots are <math>-57</math>, <math>-19</math>, <math>-3</math>, <math>-1</math>, <math>1</math>, <math>3</math>, <math>19</math>, and <math>57</math>. After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots. <math>\square</math> | '''Solution''': The polynomial has leading coefficient <math>1</math> and constant term <math>3 \cdot 19</math>, so the rational root theorem guarantees that the only possible rational roots are <math>-57</math>, <math>-19</math>, <math>-3</math>, <math>-1</math>, <math>1</math>, <math>3</math>, <math>19</math>, and <math>57</math>. After testing every number, we find that none of these are roots of the polynomial; thus, the polynomial has no rational roots. <math>\square</math> | ||
− | === | + | === Example 2 === |
''Factor the polynomial <math>x^3-5x^2+2x+8</math>.'' | ''Factor the polynomial <math>x^3-5x^2+2x+8</math>.'' | ||
− | '''Solution''': After testing the divisors of 8, we find that it has roots <math>-1</math>, <math>2</math>, and <math>4</math>. Then because it has leading coefficient <math>1</math>, the [[factor theorem]] tells us that it has the factorization <math>(x-4)(x-2)(x+1)</math>. <math>\square</math> | + | '''Solution''': After testing the divisors of 8, we find that it has roots <math>-1</math>, <math>2</math>, and <math>4</math>. Then because it has leading coefficient <math>1</math>, the [[factor theorem]] tells us that it has the factorization <math>(x-4)(x-2)(x+1), x={-1, 2, 4}</math>. <math>\square</math> |
− | === | + | === Example 3 === |
''Using the rational root theorem, prove that <math>\sqrt{2}</math> is irrational.'' | ''Using the rational root theorem, prove that <math>\sqrt{2}</math> is irrational.'' | ||
− | '''Solution''': The polynomial <math>x^2 - 2</math> has roots <math> | + | '''Solution''': The polynomial <math>x^2 - 2</math> has roots <math>\pm \sqrt{2}</math>. The rational root theorem guarantees that the only possible rational roots of this polynomial are <math>-2, -1, 1</math>, and <math>2</math>. Testing these, we find that none are roots of the polynomial, and so it has no rational roots. Then because <math>\sqrt{2}</math> is a root of the polynomial, it cannot be a rational number. <math>\square</math> |
== See also == | == See also == |
Latest revision as of 12:07, 20 February 2024
In algebra, the rational root theorem states that given an integer polynomial with leading coefficient and constant term , if has a rational root in lowest terms, then and .
This theorem is most often used to guess the roots of polynomials. It sees widespread usage in introductory and intermediate mathematics competitions.
Proof
Let be a rational root of , where every is an integer; 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.
Intro to Rational Roots theorem: https://www.youtube.com/shorts/wKpmfnyKeeM
Examples
Here are some problems with solutions that utilize the rational root theorem.
Example 1
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.
Example 2
Factor the polynomial .
Solution: After testing the divisors of 8, we find that it has roots , , and . Then because it has leading coefficient , the factor theorem tells us that it has the factorization .
Example 3
Using the rational root theorem, prove that is irrational.
Solution: The polynomial has roots . The rational root theorem guarantees that the only possible rational roots of this polynomial 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.