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The rational root theorem is a special case (for a single linear factor) of Gauss's lemma on the factorization of polynomials. The integral root theorem is the special case of the rational root theorem when the leading coefficient is a n = 1.
If =, then it says a rational root of a monic polynomial over integers is an integer (cf. the rational root theorem). To see the statement, let a / b {\displaystyle a/b} be a root of f {\displaystyle f} in F {\displaystyle F} and assume a , b {\displaystyle a,b} are relatively prime .
Abel–Ruffini theorem; Bring radical; Binomial theorem; Blossom (functional) Root of a function; nth root (radical) Surd; Square root; Methods of computing square roots; Cube root; Root of unity; Constructible number; Complex conjugate root theorem; Algebraic element; Horner scheme; Rational root theorem; Gauss's lemma (polynomial) Irreducible ...
Theorem — The number of strictly positive roots (counting multiplicity) of is equal to the number of sign changes in the coefficients of , minus a nonnegative even number. If b 0 > 0 {\displaystyle b_{0}>0} , then we can divide the polynomial by x b 0 {\displaystyle x^{b_{0}}} , which would not change its number of strictly positive roots.
Sometimes one or more roots of a polynomial are known, perhaps having been found using the rational root theorem. If one root r of a polynomial P(x) of degree n is known then polynomial long division can be used to factor P(x) into the form (x − r)Q(x) where Q(x) is a polynomial of degree n − 1. Q(x) is simply the quotient obtained from the ...
Rational root theorem; Routh–Hurwitz theorem; S. Schwartz–Zippel lemma; Sturm's theorem This page was last edited on 1 November 2020, at 19:24 (UTC). Text is ...
This results from the rational root theorem, which asserts that, if the rational number is a root of a polynomial with integer coefficients, then q is a divisor of the leading coefficient; so, if the polynomial is monic, then =, and the number is an integer.
By the rational root theorem, this has no rational zeroes. Neither does it have linear factors modulo 2 or 3. The Galois group of f(x) modulo 2 is cyclic of order 6, because f(x) modulo 2 factors into polynomials of orders 2 and 3, (x 2 + x + 1)(x 3 + x 2 + 1). f(x) modulo 3 has no linear or quadratic factor, and hence is irreducible. Thus its ...