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This can be proved as follows. First, if r is a root of a polynomial with real coefficients, then its complex conjugate is also a root. So the non-real roots, if any, occur as pairs of complex conjugate roots. As a cubic polynomial has three roots (not necessarily distinct) by the fundamental theorem of algebra, at least one root must be real.
Finding the roots (zeros) of a given polynomial has been a prominent mathematical problem.. Solving linear, quadratic, cubic and quartic equations in terms of radicals and elementary arithmetic operations on the coefficients can always be done, no matter whether the roots are rational or irrational, real or complex; there are formulas that yield the required solutions.
Vieta's formulas are frequently used with polynomials with coefficients in any integral domain R.Then, the quotients / belong to the field of fractions of R (and possibly are in R itself if happens to be invertible in R) and the roots are taken in an algebraically closed extension.
For simple roots, this results immediately from the implicit function theorem. This is true also for multiple roots, but some care is needed for the proof. A small change of coefficients may induce a dramatic change of the roots, including the change of a real root into a complex root with a rather large imaginary part (see Wilkinson's polynomial).
The solutions of the system are in one-to-one correspondence with the roots of h and the multiplicity of each root of h equals the multiplicity of the corresponding solution. The solutions of the system are obtained by substituting the roots of h in the other equations. If h does not have any multiple root then g 0 is the derivative of h.
It follows from the present theorem and the fundamental theorem of algebra that if the degree of a real polynomial is odd, it must have at least one real root. [2] This can be proved as follows. Since non-real complex roots come in conjugate pairs, there are an even number of them;
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.
By the fundamental theorem of algebra, if the monic polynomial equation x 2 + bx + c = 0 has complex coefficients, it must have two (not necessarily distinct) complex roots. Unfortunately, the discriminant b 2 − 4c is not as useful in this situation, because it may be a complex number. Still, a modified version of the general theorem can be ...