answersLogoWhite

0

1+x2 is a polynomial and doesn't have a real root.

User Avatar

Wiki User

14y ago

What else can I help you with?

Related Questions

The polynomial given below has rootss?

You forgot to copy the polynomial. However, the Fundamental Theorem of Algebra states that every polynomial has at least one root, if complex roots are allowed. If a polynomial has only real coefficients, and it it of odd degree, it will also have at least one real solution.


Can you have quadratic function with one real root and are complex root?

Yes, but in this case, the coefficients of the polynomial can not all be real.


What is the square root of a polynomial?

The square root of a polynomial is another polynomial that, when multiplied by itself, yields the original polynomial. Not all polynomials have a square root that is also a polynomial; for example, the polynomial (x^2 + 1) does not have a polynomial square root in the real number system. However, some polynomials, like (x^2 - 4), have polynomial square roots, which in this case would be (x - 2) and (x + 2). Finding the square root of a polynomial can involve techniques such as factoring or using the quadratic formula for quadratic polynomials.


The root of transcedence?

in math, a real number that is not the root of any polynomial with rational coefficients is called transcendental.


Is it true that the degree of polynomial function determine the number of real roots?

Sort of... but not entirely. Assuming the polynomial's coefficients are real, the polynomial either has as many real roots as its degree, or an even number less. Thus, a polynomial of degree 4 can have 4, 2, or 0 real roots; while a polynomial of degree 5 has either 5, 3, or 1 real roots. So, polynomial of odd degree (with real coefficients) will always have at least one real root. For a polynomial of even degree, this is not guaranteed. (In case you are interested about the reason for the rule stated above: this is related to the fact that any complex roots in such a polynomial occur in conjugate pairs; for example: if 5 + 2i is a root, then 5 - 2i is also a root.)


What does it mean to be a root of a polynomial?

A root of a polynomial is a value of the variable for which the polynomial evaluates to zero. In other words, if ( p(x) ) is a polynomial, then a number ( r ) is a root if ( p(r) = 0 ). Roots can be real or complex and are critical for understanding the behavior and graph of the polynomial function. The Fundamental Theorem of Algebra states that a polynomial of degree ( n ) has exactly ( n ) roots, counting multiplicities.


What is the number which when substituted in a polynomial makes its value zero?

A root.


How to tell if there are no real roots?

The real roots of what, exactly? If you mean a square trinomial, then: If the discriminant is positive, the polynomial has two real roots. If the discriminant is zero, the polynomial has one (double) real root. If the discriminant is negative, the polynomial has two complex roots (and of course no real roots). The discriminant is the term under the square root in the quadratic equation, in other words, b2 - 4ac.


What type of number is square root of 41?

The square root of 41 is an irrational , algebraic, real number. It is real, is the root of an polynomial equation with integer coefficients, but can not be expressed as a ratio of integers.


Is it possible to find a polynomial of degree 3 that has -2 as its only real zero?

5


Can a polynomial be no rational zeros but have real zeros?

Yes, a polynomial can have no rational zeros while still having real zeros. This occurs, for example, in the case of a polynomial like (x^2 - 2), which has real zeros ((\sqrt{2}) and (-\sqrt{2})) but no rational zeros. According to the Rational Root Theorem, any rational root must be a factor of the constant term, and if none exist among the possible candidates, the polynomial can still have irrational real roots.


What polynomial has one root that equals 2 plus i name one other root of this polynomial?

There are infinitely many polynomials which meet the requirement.The polynomial is x2 + (-a - bi - 3)x + (2a - 2bi - ai + b) = 0and the other root is x = a + bi.There is nothing in the question which requires its coefficients to be real.