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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.
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.
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 .
The rational root theorem (or integer root theorem) may be used to show that any square root of any natural number that is not a perfect square is irrational. For other proofs that the square root of any non-square natural number is irrational, see Quadratic irrational number or Infinite descent.
An element a of F is integral over R if it is a root of a monic polynomial with coefficients in R. A complex number that is integral over the integers is called an algebraic integer. This terminology is motivated by the fact that the integers are exactly the rational numbers that are also algebraic integers.
Triangles whose side lengths and areas are all rational numbers (positive rational solutions of the above equation) are sometimes also called Heronian triangles or rational triangles; [5] in this article, these more general triangles will be called rational Heronian triangles. Every (integral) Heronian triangle is a rational Heronian triangle.
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 ...
Since x = cos 20° is a root of p(t), the minimal polynomial for cos 20° is a factor of p(t). Because p(t) has degree 3, if it is reducible over by Q then it has a rational root. By the rational root theorem, this root must be ±1, ± 1 / 2 , ± 1 / 4 or ± 1 / 8 , but none of these is a root.