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What are the roots of polynomial?
The "roots" of a polynomial are the solutions of the equation polynomial = 0. That is, any value which you can replace for "x", to make the polynomial equal to zero.
Can a polynomial equation have no answer?
It depends on the domain. In the complex domain, a polynomial of order n must have n solutions, although some of these may be multiple solutions. In the real domain, a polynomial of odd order must have at least one real solution, while a polynomial of even order may have no real solutions.
How can you determine whether a polynomial equation has imaginary solutions?
To determine whether a polynomial equation has imaginary solutions, you must first identify what type of equation it is. If it is a quadratic equation, you can use the quadratic formula to solve for the solutions. If the equation is a cubic or higher order polynomial, you can use the Rational Root Theorem to determine if there are any imaginary solutions. The Rational Root Theorem states that if a polynomial equation has rational solutions, they must be a factor of the constant term divided by a factor of the leading coefficient. If there are no rational solutions, then the equation has imaginary solutions. To use the Rational Root Theorem, first list out all the possible rational solutions. Then, plug each possible rational solution into the equation and see if it is a solution. If there are any solutions, then the equation has imaginary solutions. If not, then there are no imaginary solutions.
What is the maximum number of solutions a quartic function can have?
I assume you mean a polynomial of degree 4. In general, a polynomial of degree "n" can be separated into "n" linear factors. As a result, a polynomial of degree "n" has exactly "n" solutions - unless two or more of the factors are repeated; in which case the corresponding solution is said to be a multiple solution. As an example: (x - 2)(x - 5) = 0 has two solutions, namely 2 and 5; while (x - 3)(x - 3)(x + 5) = 0 is of degree three, but has only two solutions, since the solution "3" is repeated. An equation such as x2 + 1 = 0 cannot be factored in the real numbers, so if you insist that the solutions be real, there are zero solutions. However, the polynomial can be factored in the complex numbers; in this case: (x + i)(x - i) = 0, resulting in the two complex solutions, -i and +i.
How do you factor a polynomial using distributive property when it has 5 terms?
Factoring a polynomial with 5 or more terms is very hard and in general impossible using only algebraic numbers. The best strategy here is to guess some 'obvious' solutions and reduce to a fourth or lower order polynomial.
