There are infinitely many polynomials which meet the requirement.The polynomial is x2 + (-a - bi - 3)x + (2a - 2bi - ai + b) = 0
and the other root is x = a + bi.
There is nothing in the question which requires its coefficients to be real.
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.
The degree of this polynomial is 2.
yes, and it is 14x
2048
y = x2 - 4x + 4 can be factored into y = (x-2)(x-2) The repeated factor is 2.
4-17i
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.
F(x) = 15x2 - 2.5 + 3 That's a quadratic or 2nd degree polynomial in x.
what kind of polynomial is shown 3x3+x+1
7X^3 Third degree polynomial.
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.
Fibonacci numbers
Both - a polynomial expression, if you like.
The degree of this polynomial is 2.