Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions

Question

n=4; -2,5,3 3+2i are zeros if (1)=-96

Answer 1

OK. If the zeros are 2, 5, 3, and 3+2i when n=4 there are no polynomials with real coefficients that have those zeros. So I'm going to assume you meant the zeros are 2, 5, and 3+2i and the extra 3 is a typo. With real coefficients, the conjugate root theorem tells us that if 3+2i is a zero so is its conjugate 3-2i. For n=4 you must have four zeros and we have them: 2, 5, 3+2i, 3-2i. Use the factored form of a polynomial:

f(x) = a·(x-p)(x-q)(x-r)(x-s)

where p, q, r, and s are the zeros:

f(x) = a·(x-2)(x-3)(x-(3+2i))(x-(3-2i))

f(x) = a·(x-2)(x-3)(x-3-2i))(x-3+2i)

To find the value of the constant a, plug in the given point (1,-96) and solve for a.