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If the discriminant of such a polynomial is negative, then both roots of the quadratic equation have imaginary parts. In particular, if b and c are real numbers and b 2 − 4 c < 0, all the convergents of this continued fraction "solution" will be real numbers, and they cannot possibly converge to a root of the form u + iv (where v ≠ 0 ...
A necessary (but not sufficient) condition for solvability is that n is not divisible by 4 or by a prime of form 4k + 3. [note 3] Thus, for example, x 2 − 3y 2 = −1 is never solvable, but x 2 − 5y 2 = −1 may be. [27] The first few numbers n for which x 2 − ny 2 = −1 is solvable are
In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial.According to the theorem, the power (+) expands into a polynomial with terms of the form , where the exponents and are nonnegative integers satisfying + = and the coefficient of each term is a specific positive integer ...
Let () be a polynomial equation, where P is a univariate polynomial of degree n. If one divides all coefficients of P by its leading coefficient, one obtains a new polynomial equation that has the same solutions and consists to equate to zero a monic polynomial. For example, the equation
[2] [3] In the 1970s Askold Khovanskii developed the theory of fewnomials that generalises Descartes' rule. [4] The rule of signs can be thought of as stating that the number of real roots of a polynomial is dependent on the polynomial's complexity, and that this complexity is proportional to the number of monomials it has, not its degree.
[17] [18] For example, the fraction 1/(x 2 + 1) is not a polynomial, and it cannot be written as a finite sum of powers of the variable x. For polynomials in one variable, there is a notion of Euclidean division of polynomials, generalizing the Euclidean division of integers.
The unique pair of values a, b satisfying the first two equations is (a, b) = (1, 1); since these values also satisfy the third equation, there do in fact exist a, b such that a times the original first equation plus b times the original second equation equals the original third equation; we conclude that the third equation is linearly ...
For each k, the polynomial () can be characterized as the unique degree k polynomial p(t) satisfying p(0) = p(1) = ⋯ = p(k − 1) = 0 and p(k) = 1. Its coefficients are expressible in terms of Stirling numbers of the first kind :
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