Solve each problem. Find a polynomial function ƒ of degree 3 with -2, 1, and 4 as zeros, and ƒ(2)=16.

Verified step by step guidance

1. Start by using the fact that the zeros of the polynomial are -2, 1, and 4. This means the polynomial can be expressed as: \( f(x) = a(x + 2)(x - 1)(x - 4) \), where \( a \) is a constant.
2. To find the value of \( a \), use the condition \( f(2) = 16 \). Substitute \( x = 2 \) into the polynomial: \( f(2) = a(2 + 2)(2 - 1)(2 - 4) \).
3. Simplify the expression: \( f(2) = a(4)(1)(-2) \).
4. Set the expression equal to 16: \( 16 = a(4)(1)(-2) \).
5. Solve for \( a \) by isolating it on one side of the equation.

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 constants.

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 zeros. Given the zeros -2, 1, and 4, the polynomial can be expressed in factored form as ƒ(x) = a(x + 2)(x - 1)(x - 4), where 'a' is a constant that can be determined using additional conditions.

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 find the constant 'a' by using the condition ƒ(2) = 16. This means substituting x = 2 into the polynomial and setting the equation equal to 16, allowing us to solve for 'a' and fully define the polynomial function.