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Moreover, if one sets x = 1 + t, one gets without computation that () = (+) is a polynomial in t with the same first coefficient 3 and constant term 1. [2] The rational root theorem implies thus that a rational root of Q must belong to {,}, and thus that the rational roots of P satisfy = + {,,,}.
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.
In mathematics, the root test is a criterion for the convergence (a convergence test) of an infinite series.It depends on the quantity | |, where are the terms of the series, and states that the series converges absolutely if this quantity is less than one, but diverges if it is greater than one.
The rational root test allows finding q and p by examining a finite number of cases (because q must be a divisor of a, and p must be a divisor of d). Thus, one root is =, and the other roots are the roots of the other factor, which can be found by polynomial long division.
This polynomial has no rational roots, since the rational root theorem shows that the only possibilities are ±1, but x 0 is greater than 1. So x 0 is an irrational algebraic number. There are countably many algebraic numbers, since there are countably many integer polynomials.
The solution in radicals (without trigonometric functions) of a general cubic equation, when all three of its roots are real numbers, contains the square roots of negative numbers, a situation that cannot be rectified by factoring aided by the rational root test, if the cubic is irreducible; this is the so-called casus irreducibilis ...
Obreschkoff–Ostrowski theorem: in blue and yellow, the regions of the complex plane where there should be no root for having 0 or 1 sign variation; on the left the regions excluded for the roots of p, on the right, the regions excluded for the roots of the transformed polynomial q; in blue, the regions that are excluded for having one sign ...
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.