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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 complex conjugate root theorem states that if the coefficients of a polynomial are real, then the non-real roots appear in pairs of the form (a + ib, a – ib). It follows that the roots of a polynomial with real coefficients are mirror-symmetric with respect to the real axis.
Root (or zero) of a polynomial: Given a polynomial p(x), the x values that satisfy p(x) = 0 are called roots (or zeroes) of the polynomial p. Graphing End behaviour –
Typically, R is the ring of the integers, the field of fractions is the field of the rational numbers and the algebraically closed field is the field of the complex numbers. Vieta's formulas are then useful because they provide relations between the roots without having to compute them.
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.
(±1 cannot be a root because a 0 /a n is not integral.) It so happens that neither 1 nor -1 is a root, but this is not implied by the rational root theorem. There is also no theorem saying that ±1 can only be a root if a 0 /a n is integral. For example, 3x 3-x 2-x-1 = 0 has a root x=1, even though -1/3 is not integral.
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 ...