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.)
5
No,
Each distinct real root is an x-intercept. So the answer is 4.
Real, complex, quaternion, etc. But more specifically it's a transcendental real number, not being the root of an integer polynomial (and certainly not the ratio of two integers).
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.
Yes, but in this case, the coefficients of the polynomial can not all be real.
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.)
in math, a real number that is not the root of any polynomial with rational coefficients is called transcendental.
A root.
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.
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.
5
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.
If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.If you mean a math problem, "root" is another word for "solution".The "root" of a polynomial in "x" is any value for "x" which will set the polynomial equal to zero, when evaluated.
There is not enough information. You can't calculate one root on the basis of another root. HOWEVER, if we assume that all the polynomial's coefficients are real, then if the polynomial has a complex root, then the complex conjugate of that root (in this case, 4 - 17i) must also be a root.
Yes, every real number has a unique cube root.