Question

Using complete sentences, describe how you would analyze the zeros of the polynomial function f(x) = 7x5+ 15x4– x3+ 4x2– 6x – 11 using Descartes’ Rule of Signs.

Answer 1

Since this is a fifth degree polynomial there are 5 zeros. Using Decarte's rule of signs, the number of variations of signs in f(x) is 3. So there are either 3 positive real zeros or 1 positive real zero.

Evaluating for f(-x) you have the function: \[-7x ^{5}+15x ^{4}+x ^{3}+4x ^{2}+6x-11\]

Using Decarte's rule of signs, the number of variations of signs in f(-x) is 2. So there are either 2 negative real zeros or zero negative zeros.

Complex zeros come in pairs. So the possible combinations of zeros are 3 positive real zeros and 2 negative real zeros; 3 positive real zeros and 2 complex zeros; or 1 positive real zero, 2 negative real zeros, and 2 complex zeros;