Difference between revisions of "Rational root theorem"
Etmetalakret (talk | contribs) |
Etmetalakret (talk | contribs) |
||
Line 9: | Line 9: | ||
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. The answers can be found [[Rational root theorem/problems | here]]. | ||
− | === | + | === Problem 1 === |
− | + | ''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>, its factorization is <math>(x-4)(x-2)(x+1)</math>. <math>\square</math> | |
− | |||
− | |||
− | === | + | === Problem 2 === |
− | + | ''Find all rational roots of the polynomial <math>x^4-x^3-x^2+x+57</math>. | |
− | + | ||
− | 3. | + | '''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> |
+ | |||
+ | === Problem 3 === | ||
+ | ''Using the rational root theorem, prove that <math>\sqrt{2}</math> is irrational.'' | ||
+ | |||
+ | '''Solution''': The polynomial <math>x^2 - 2</math> has roots <math>-\sqrt{2}</math> and <math>\sqrt{2}</math>. The rational root theorem garuntees that the only possible rational roots 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 == |
Revision as of 13:16, 15 July 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. The answers can be found here.
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.