Final answer:
The polynomial p(x) = x³ – 3x² + 2x – 6 likely has only one real root. This is based on applying the Rational Root Theorem and checking the number of sign changes using Descartes' Rule of Signs.
Step-by-step explanation:
Given the polynomial function p(x) = x³ – 3x² + 2x – 6, we'll explore whether it has only one real zero or root. By applying the Rational Root Theorem, we can list the potential rational roots of the polynomial, which are the factors of the constant term (-6) divided by the factors of the leading coefficient (1 in this case).
However, to confirm whether the polynomial indeed has only one real zero, we would have to apply the Descartes' Rule of Signs. This rule considers the number of sign changes in the polynomial. Each sign change indicates a possible real root. In our polynomial, there are two sign changes (from x³ to - 3x² and from 2x to -6), suggesting that there could be two or zero positive real roots. When we substitute -x into the polynomial, no sign changes occur, so there are no negative real roots.
However, this is a cubic function, it must have at least one real root. Since Descartes' Rule of Signs doesn't confirm the existence of two positive roots and there are no negative roots, we can conclude that there is only one real root. Though, without further algebraic or numerical verification, we can't identify the specific value of the root.
Learn more about Polynomial Roots