Find a polynomial function ƒ(x) of degree 3 with real coefficients that satisfies the given conditions. See Example 4. Zeros of -3, 1, and 4; ƒ(2)=30

Verified step by step guidance

1. Identify the zeros of the polynomial: -3, 1, and 4.
2. Write the polynomial in factored form using the zeros: f(x) = a(x + 3)(x - 1)(x - 4).
3. Use the given condition f(2) = 30 to find the value of 'a'.
4. Substitute x = 2 into the polynomial: f(2) = a(2 + 3)(2 - 1)(2 - 4).
5. Solve for 'a' by setting the equation equal to 30 and simplifying: 30 = a(5)(1)(-2).

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. In this case, a degree 3 polynomial will have 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 problem, the given zeros are -3, 1, and 4, which means the polynomial can be expressed in factored form as ƒ(x) = a(x + 3)(x - 1)(x - 4), where 'a' is a constant that scales the polynomial.

Evaluating Polynomial Functions

Evaluating a polynomial function involves substituting a specific value for the variable and calculating the result. In this case, we need to ensure that the polynomial satisfies the condition ƒ(2) = 30. This means substituting x = 2 into the polynomial and solving for the coefficient 'a' to ensure the output equals 30, which helps in determining the specific polynomial that meets all given conditions.