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It depends on the polynomial and your degree of sophistication. In the complex domain, it will have six solutions, although not all of them need be different. If the coefficients are all real, then it will have 0, 2, 4 or 6 real solutions in the real domain.
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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.
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.
What is the discriminant of the polynomial below?
That depends on the values of the polynomial but in general:- If the discriminant is greater than zero it has 2 solutions If the discriminant is equal to zero then it has 2 equal solutions If the discriminant is less than zero it has no solutions
Is it possible to have imaginary solutions when solving a polynomial equation?
yes
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 point at which a graph crosses the x-axis?
For a line, this is the x-intercept. For a polynomial, these points are the roots or solutions of the polynomial at which y=0.
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 can you find out how many solutions an equation has?
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.
How many solutions does an polynomial equation have?
Depends on degree of highest term. a^3 + bX^2 + cX + d = 0 has three solutions. And so on. Finding them is another matter.
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.
Number of solutions of a quadratic polynomial?
If you allow solutions that are complex numbers, then there are always two solutions, although the two solutions may be coincident so as to appear as a single solution. If limited to real solution, then two or none, although again, a coincident pair of solutions will appear as a single solution.
What are the solutions of the equation x3 plus 3x2-x-3 equals 0?
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.
