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What would be a counterexample for this statement A polynomial equation of degree three always has three real solutions?
Wiki User
∙ 11y ago
A cubic polynomial of the form
y = (x - d)*(ax2+ bx + c) where
a, b, c and d are real constants and b2-4ac < 0
has only one root, which is x = d.
Wiki User
∙ 11y ago
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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 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 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.
According to the Zero Product Rule the solutions to the simpler equations are also the solutions to the original equation. A.True B.False?
The statement is true.
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 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.
Is it possible to have imaginary solutions when solving a polynomial equation?
yes
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 statement must be true of an equation before you can use the quadratic formula to find the solutions?
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.
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.
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.
The set of values that makes a statement true?
If the statement is a mathematical equation, than those values are its "solutions". The number of them depends on the equation. There may be only one, more than one, or no solutions at all.
Who discovered polynomial?
In the 1880s, Poincaré created functions which give the solution to the order polynomial equation to the order of the polynomial equation
What is the difference between a polynomial and a quadratic equation?
A quadratic equation is of degree 2, that is, the highest power is 2. A polynomial is not an equation, however, you can convert it into an equation by setting the polynomial equal to zero for example. A polynomial EQUATION can be of any degree: 1, 2, 3, 4, etc.
What is the difference in evaluating a polynomial and solving a polynomial?
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.
How do you answer polynomial?
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.
