If a polynomial function f(x) has roots 4 - 13i and 5, what must be a factor of f(x)?

Solution:

If a quadratic equation has complex roots then it will be in pairs. If a polynomial of degree three has complex roots then they will be in pairs. So if it has one root as a + bi then the second root has to be a - bi. The third root will be the real root because a polynomial of any odd degree will have at least one real root.

In the given problem it can be stated that there is at least one real root and the factor formed by that root is (x - 5).To determine the number of positive roots of a polynomial equation there is a rule which is known as Descartes’ Rule of Signs. It states:

“The number of positive roots is at most equal to the number of sign changes. It can also be less than that by 2, 4, … i.e. if there are k sign changes in f(x), the number of positive roots could be k, k-2, and k-4,.....”

The above polynomial is at least of degree 3 because one of the roots is a complex root 4 - 13i and complex roots occur in pairs. The real root of the polynomial is 5 and hence one of the factors of the polynomial must be: (x - 5)

Thus the factors of the given polynomial f(x) are (x-5)(x+4 - 13i)(x -4 + 13i)

If a polynomial function f(x) has roots 4 - 13i and 5, what must be a factor of f(x)?

Summary:

If a polynomial function f(x) has roots 4 - 13i and 5, then factors are  (x-5)(x+4 - 13i)(x -4 + 13i).