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Ch. 3 - Polynomial and Rational Functions
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 4, Problem 37

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

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Start by writing the general form of a cubic polynomial function with the given zeros. Since the zeros are -2, 1, and 4, the function can be expressed as: \[f(x) = a(x + 2)(x - 1)(x - 4)\] where \(a\) is a constant coefficient to be determined.
Expand the factors \((x + 2)(x - 1)(x - 4)\) step-by-step. First, multiply two of the binomials, for example \((x + 2)(x - 1)\), and then multiply the result by the third binomial \((x - 4)\).
After expanding, you will have a cubic polynomial in the form \(a(x^3 + bx^2 + cx + d)\). Keep the coefficient \(a\) as a variable since it is not yet known.
Use the given condition \(f(2) = 16\) to find the value of \(a\). Substitute \(x = 2\) into the expanded polynomial and set the expression equal to 16. This will give you an equation in terms of \(a\).
Solve the equation for \(a\), then write the final polynomial function \(f(x)\) by substituting the value of \(a\) back into the expanded polynomial.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Polynomial Functions and Degree

A polynomial function is an expression involving variables raised to whole-number exponents and coefficients. The degree of a polynomial is the highest exponent of the variable, which determines the function's general shape and number of roots. For a degree 3 polynomial, there are up to three zeros or roots.
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Zeros (Roots) of a Polynomial

Zeros of a polynomial are the values of the variable that make the function equal to zero. If -2, 1, and 4 are zeros, then (x + 2), (x - 1), and (x - 4) are factors of the polynomial. The polynomial can be expressed as a product of these factors multiplied by a constant.
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Using a Point to Find the Leading Coefficient

Given a point on the polynomial, such as ƒ(2) = 16, you can substitute x = 2 into the factored form to solve for the leading coefficient. This constant scales the polynomial so it passes through the given point, ensuring the function satisfies all conditions.
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