Answer:
To construct a fourth-degree polynomial with real integer coefficients having the roots √3 and 1 - i, we can use the fact that the complex conjugate of 1 - i is 1 + i, which is also a root of the polynomial. Therefore, we have:
(x - √3)(x - 1 + i)(x - 1 - i)(x - (1 + i)) = 0
Expanding this expression gives:
(x - √3)(x^3 - 2x^2 + 3x - 3)(x^2 - 2x + 2) = 0
Multiplying out the factors gives:
x^6 - 2x^5 + (5 - √3)x^4 - (8 + 2√3)x^3 + (6 + 4√3)x^2 - (6√3 + 12)x + 6√3 = 0
So the required fourth-degree polynomial with real integer coefficients is:
x^4 - (5 - √3)x^3 + (8 + 2√3)x^2 - (6 + 4√3)x + 6√3 = 0
To determine the possible numbers of imaginary roots a sixth-degree function can have, we can use the Fundamental Theorem of Algebra, which states that a polynomial of degree n has exactly n complex roots, counting multiplicities. Therefore, a sixth-degree function can have at most six complex roots. Since the coefficients are real, complex roots must come in conjugate pairs, so the function can have at most three distinct imaginary roots. However, it is also possible that the function has no imaginary roots or one imaginary root (with multiplicity), in which case the remaining roots are real.
Explanation: