Force of Wind The force of the wind blowing on a vertical surface varies jointly as the area of the surface and the square of the velocity. If a wind of 40 mph exerts a force of 50 lb on a surface of 1/2 ft2, how much force will a wind of 80 mph place on a surface of 2 ft2?
Ch. 3 - Polynomial and Rational Functions

Chapter 4, Problem 41
For each polynomial function, use the remainder theorem to find ƒ(k). ƒ(x) = x2 + 4; k = 2i
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Recall the Remainder Theorem, which states that the remainder when a polynomial ƒ(x) is divided by (x - k) is equal to ƒ(k). So, to find ƒ(k), we simply substitute k into the polynomial ƒ(x).
Given the polynomial function ƒ(x) = x^2 + 4, and k = 2i, substitute x with 2i in the polynomial: ƒ(2i) = (2i)^2 + 4.
Calculate the square of 2i: (2i)^2 = 2^2 * i^2 = 4 * i^2.
Recall that i is the imaginary unit where i^2 = -1, so replace i^2 with -1 in the expression: 4 * (-1) = -4.
Now, add the constant term: ƒ(2i) = -4 + 4. This expression represents the value of ƒ(k) using the Remainder Theorem.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Polynomial Functions
A polynomial function is an expression consisting of variables and coefficients combined using addition, subtraction, multiplication, and non-negative integer exponents. Understanding the structure of polynomial functions helps in evaluating them at specific values, including complex numbers.
Recommended video:
Introduction to Polynomial Functions
Remainder Theorem
The Remainder Theorem states that when a polynomial ƒ(x) is divided by (x - k), the remainder is equal to ƒ(k). This allows us to find the value of the polynomial at k by simply evaluating ƒ(k), without performing full polynomial division.
Recommended video:
Higher Powers of i
Complex Numbers
Complex numbers include a real part and an imaginary part, expressed as a + bi, where i is the imaginary unit with i² = -1. Evaluating polynomials at complex values requires substituting and simplifying using the properties of i.
Recommended video:
Dividing Complex Numbers
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