Question

Let P(x) be a polynomial with real coefficients and let b>0 . Use the Division Algorithm to write P(x)=(x-b) • Q(x)+r . Suppose that r ≥ 0 and that all the coefficients in Q(x) are nonnegative. Let z > b (a) Show that P(z)>0 .

Answer 1

If z > b and P(x)=(x-b) • Q(x)+r, with r ≥ 0 and nonnegative coefficients in Q(x), then P(z) > 0.

Step-by-step explanation:

To show that P(z) > 0, we can use the Division Algorithm. According to the question, P(x) can be written as P(x)=(x-b) • Q(x)+r, where r ≥ 0 and all coefficients in Q(x) are nonnegative. Let's assume that z > b. Since r ≥ 0, the remainder term in the Division Algorithm is nonnegative for any value of x. Therefore, when we substitute z into the equation P(x)=(x-b) • Q(x)+r, we have P(z)=(z-b) • Q(z)+r, where r ≥ 0. Since z > b, (z-b) is positive. Additionally, we know that all coefficients in Q(x) are nonnegative, so (z-b) • Q(z) will be nonnegative as well. Adding these two nonnegative terms to the nonnegative remainder r, we can conclude that P(z) will be greater than 0.