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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.
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.
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 square root of 2 is equal to the length of the hypotenuse of a right triangle with legs of length 1 and is therefore a constructible number. In geometry and algebra, a real number is constructible if and only if, given a line segment of unit length, a line segment of length | | can be constructed with compass and straightedge in a finite number of steps.
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 ...
Casus irreducibilis occurs when none of the roots are rational and when all three roots are distinct and real; the case of three distinct real roots occurs if and only if q 2 / 4 + p 3 / 27 < 0, in which case Cardano's formula involves first taking the square root of a negative number, which is imaginary, and then taking the ...
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.