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A solution of a polynomial system is a tuple of values of (x 1, ..., x m) that satisfies all equations of the polynomial system. The solutions are sought in the complex numbers, or more generally in an algebraically closed field containing the coefficients. In particular, in characteristic zero, all complex solutions are sought. Searching for ...
For these problems, it is very easy to tell whether solutions exist, but thought to be very hard to tell how many. Many of these problems are #P-complete, and hence among the hardest problems in #P, since a polynomial time solution to any of them would allow a polynomial time solution to all other #P problems.
Finding roots −1/2, −1/ √ 2, and 1/ √ 2 of the cubic 4x 3 + 2x 2 − 2x − 1, showing how negative coefficients and extended segments are handled. Each number shown on a colored line is the negative of its slope and hence a real root of the polynomial. To employ the method, a diagram is drawn starting at the origin.
This polynomial is further reduced to = + + which is shown in blue and yields a zero of −5. The final root of the original polynomial may be found by either using the final zero as an initial guess for Newton's method, or by reducing () and solving the linear equation. As can be seen, the expected roots of −8, −5, −3, 2, 3, and 7 were ...
A solution in radicals or algebraic solution is an expression of a solution of a polynomial equation that is algebraic, that is, relies only on addition, subtraction, multiplication, division, raising to integer powers, and extraction of n th roots (square roots, cube roots, etc.). A well-known example is the quadratic formula
In elementary algebra, completing the square is a technique for converting a quadratic polynomial of the form + + to the form + for some values of and . [1] In terms of a new quantity x − h {\displaystyle x-h} , this expression is a quadratic polynomial with no linear term.
Instantiating a symbolic solution with specific numbers gives a numerical solution; for example, a = 0 gives (x, y) = (1, 0) (that is, x = 1, y = 0), and a = 1 gives (x, y) = (2, 1). The distinction between known variables and unknown variables is generally made in the statement of the problem, by phrases such as "an equation in x and y ", or ...
An n-variable instance of 3-SAT can be realized as a positivity problem on a polynomial with n variables and d=4. This proves that positivity testing is NP-Hard . More precisely, assuming the exponential time hypothesis to be true, v ( n , d ) = 2 Ω ( n ) {\displaystyle v(n,d)=2^{\Omega (n)}} .
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