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Bézout's theorem asserts that a well-behaved system whose equations have degrees d 1, ..., d n has at most d 1 ⋅⋅⋅d n solutions. This bound is sharp. If all the degrees are equal to d, this bound becomes d n and is exponential in the number of variables. (The fundamental theorem of algebra is the special case n = 1.)
If w 1, w 2 and w 3 are the three cube roots of W, then the roots of the original depressed cubic are w 1 − p / 3w 1 , w 2 − p / 3w 2 , and w 3 − p / 3w 3 . The other root of the quadratic equation is − p 3 27 W . {\displaystyle \textstyle -{\frac {p^{3}}{27W}}.}
For polynomials in one variable, there is a notion of Euclidean division of polynomials, generalizing the Euclidean division of integers. [e] This notion of the division a(x)/b(x) results in two polynomials, a quotient q(x) and a remainder r(x), such that a = b q + r and degree(r) < degree(b).
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
In mathematics, Ruffini's rule is a method for computation of the Euclidean division of a polynomial by a binomial of the form x – r. It was described by Paolo Ruffini in 1809. [1] The rule is a special case of synthetic division in which the divisor is a linear factor.
Hilbert's tenth problem is the tenth on the list of mathematical problems that the German mathematician David Hilbert posed in 1900. It is the challenge to provide a general algorithm that, for any given Diophantine equation (a polynomial equation with integer coefficients and a finite number of unknowns), can decide whether the equation has a solution with all unknowns taking integer values.
Divide the highest term of the remainder by the highest term of the divisor (3x ÷ x = 3). Place the result (+3) below the bar. 3x has been divided leaving no remainder, and can therefore be marked as used. The result 3 is then multiplied by the second term in the divisor −3 = −9. Determine the partial remainder by subtracting −4 − (− ...
And, substitution allows one to derive restrictions on the possible values, or show what conditions the statement holds under. For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1.