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.
Not that it matters anymore since this was posted so long ago.. but the answer is 4-17i. Not sure why exactly. I just decided that maybe if I were to change the sign to negative, maybe it would work, which it did! Yes, I am smart. No need to thank me.
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.
what kind of polynomial is shown 3x3+x+1
7X^3 Third degree polynomial.
F(x) = 15x2 - 2.5 + 3 That's a quadratic or 2nd degree polynomial in x.
Both - a polynomial expression, if you like.
The degree of this polynomial is 2.
Fibonacci numbers