Find a polynomial function ƒ(x) of degree 3 with real coefficients that satisfies the given conditions. See Example 4. Zeros of -2, 1, and 0; ƒ(-1)=-1

Verified step by step guidance

1. Identify the zeros of the polynomial: -2, 1, and 0. Since the polynomial is of degree 3, it can be expressed as \( f(x) = a(x + 2)(x - 1)x \).
2. Expand the polynomial expression: First, multiply \((x + 2)(x - 1)\) to get \(x^2 + x - 2\).
3. Multiply the result by \(x\) to get \(x(x^2 + x - 2) = x^3 + x^2 - 2x\).
4. Include the leading coefficient \(a\) to form the polynomial: \(f(x) = a(x^3 + x^2 - 2x)\).
5. Use the condition \(f(-1) = -1\) to find \(a\): Substitute \(-1\) into the polynomial and solve for \(a\) using \(a((-1)^3 + (-1)^2 - 2(-1)) = -1\).

Key Concepts

Polynomial Functions

A polynomial function is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. The degree of a polynomial is determined by the highest power of the variable. For example, a polynomial of degree 3 can be expressed in the form ƒ(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are real coefficients.

Zeros of a Polynomial

The zeros (or roots) of a polynomial are the values of x for which the polynomial evaluates to zero. For a polynomial of degree 3, there can be up to three real zeros. In this case, the given zeros are -2, 1, and 0, which means the polynomial can be expressed as ƒ(x) = k(x + 2)(x - 1)(x), where k is a non-zero constant.

Evaluating Polynomial Functions

Evaluating a polynomial function involves substituting a specific value for the variable and calculating the result. In this problem, we need to ensure that the polynomial satisfies the condition ƒ(-1) = -1. This means substituting -1 into the polynomial and solving for the constant k to meet this requirement, ensuring the function meets all specified conditions.