An identity equation has infinite solutions.
In the 1880s, Poincaré created functions which give the solution to the order polynomial equation to the order of the polynomial equation
Evaluating a polynomial is finding the value of the polynomial for a given value of the variable, usually denoted by x. Solving a polynomial equation is finding the value of the variable, x, for which the polynomial equation is true.
The quadratic formula can be used to find the solutions of a quadratic equation - not a linear or cubic, or non-polynomial equation. The quadratic formula will always provide the solutions to a quadratic equation - whether the solutions are rational, real or complex numbers.
By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.By solving it. There is no single easy way to solve all equations; different types of equations required different methods. You have to learn separately how to solve equations with integer polynomials, rational equations (where polynomials can also appear in the denominator), equations with square roots and other roots, trigonometric equations, and others.Sometimes, the knowledge of a type of equations can help you quickly guess the number of solutions. Here are a few examples. An equation like:sin(x) = 0.5has an infinite number of solutions, because the sine function is periodic. An equation with a polynomial - well, in theory, you can factor a polynomial of degree "n" into "n" linear factors, meaning the polynomial can have "n" solutions. However, it may have multiple solutions, that is, some of the factors may be equal. Also, some of the solutions may be complex. A real polynomial of odd degree has at least one real solution.
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.
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.
yes
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.
No. A polynomial can have as many degrees as you like.
the solutions to this equation are -1,+1 and -3. you can solve this equation by using the polynomial long division method. we basically want to factorize this and polynomial and equate its factors to zero and obtain the roots of the equation. By hit and trial , it clear that x=1 i.e is a root of this equation. So (x-1) should be a factor of the given polynomial (LHS). Divide the polynomial by x-1 using long division method and you will get the quotient as x2+4x+3 and remainder would be 0 ( it should be 0 as we are dividing the polynomial with its factor. Eg when 8 is divided by any of its factor like 4,2 .. remainder is always zero ) Now, we can write the given polynomial as product of its factors as x3+3x2-x-3 = (x-1)(x2+4x+3) =(x-1)(x+1)(x+3) [by splitting middle term method] so the solutions for the given polynomial are obtained when RHS = 0, Hence x=-1 , X = +1, x=-3 are the solutions for this equation.
An identity equation has infinite solutions.
In the 1880s, Poincaré created functions which give the solution to the order polynomial equation to the order of the polynomial equation
It will depend on the equation.
Evaluating a polynomial is finding the value of the polynomial for a given value of the variable, usually denoted by x. Solving a polynomial equation is finding the value of the variable, x, for which the polynomial equation is true.
You can evaluate a polynomial, you can factorise a polynomial, you can solve a polynomial equation. But a polynomial is not a specific question so it cannot be answered.
There is no simple method. The answer depends partly on the variable's domain. For example, 2x = 3 has no solution is x must be an integer, or y^2 = -9 has no solution if y must be a real number but if it can be a complex number, it has 2 solutions.