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.
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.
The degree of this polynomial is 2.
yes, and it is 14x
2048
2 is.
4-17i
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.
False
A fifth degree polynomial.
The actual equation itself is the polynomial. There is no polynomial for it, and your question doesn't really make sense.
It is a quadratic polynomial.
7X^3 Third degree polynomial.
what kind of polynomial is shown 3x3+x+1
F(x) = 15x2 - 2.5 + 3 That's a quadratic or 2nd degree polynomial in x.
The degree of this polynomial is 2.
Both - a polynomial expression, if you like.
Fibonacci numbers