Problem 33 - Precalculus with Calculus Previews

Question

In Problems \(33-40\), use synthetic division and the Remainder Theorem to find \(f(c)\) for the given value of c. $$ f(x)=4 x^{2}-2 x+9 ; c=-3 $$

Step-by-Step Solution

Answer

\( f(-3) = 51 \)

Step 1: Set up synthetic division

We need to find \( f(-3) \) using synthetic division. In synthetic division, we write the constant \( c \) on the left and the coefficients of the polynomial \( f(x) = 4x^2 - 2x + 9 \) in descending order to the right. So we have \( -3 \) on the left, and the coefficients \( [4, -2, 9] \) on the right.

Step 2: Bring down the first coefficient

Bring down the first coefficient, which is 4, into the row below the line. This is our starting point for the calculations.

Step 3: Multiply and add

Multiply the value brought down (4) by \( c = -3 \) and add the result to the next coefficient. Therefore, \( 4 \times -3 = -12 \), and \(-12 + (-2) = -14 \). Write \(-14 \) below \(-2 \).

Step 4: Repeat the process

Now take \(-14 \) and multiply it by \( c = -3 \). This gives \(-14 \times -3 = 42 \). Add \( 42 \) to the last coefficient (9): \( 42 + 9 = 51 \). Write \( 51 \) as the last number in the bottom row.

Step 5: Interpret the result

The last number, \( 51 \), is the remainder. According to the Remainder Theorem, this remainder is \( f(-3) \). Therefore, \( f(-3) = 51 \).

Key Concepts

Remainder Theorem

The Remainder Theorem is a useful tool in algebra that connects division of polynomials and evaluation of functions. It states that for any polynomial \( f(x) \), when you divide \( f(x) \) by \( x-c \), the remainder of this division is \( f(c) \). This means that if you want to find the value of a polynomial at a specific point, \( c \), you can perform synthetic division and the final remainder will be the value of the polynomial at \( c \).

For example, in our given problem, we used the Remainder Theorem to find \( f(-3) \). By applying synthetic division, we were able to directly find out that \( f(-3) = 51 \) without having to substitute \( -3 \) into the polynomial and calculate. This method is both quick and efficient, especially when dealing with higher degree polynomials.

Polynomials

A polynomial is a mathematical expression consisting of variables, coefficients, and the operations of addition, subtraction, and multiplication. The degree of a polynomial is the highest power of the variable in the expression. Polynomials are named according to their degree:

• Linear polynomial: Degree 1, e.g., \( ax + b \)
• Quadratic polynomial: Degree 2, e.g., \( ax^2 + bx + c \)
• Cubic polynomial: Degree 3, and so on.

In this exercise, our polynomial \( 4x^2 - 2x + 9 \) is a quadratic polynomial since its highest power is 2. Polynomials have several important properties, which make them central in algebra, calculus, and many other areas of mathematics. They are also foundational for understanding complex functions and higher mathematics. Understanding polynomials and their behavior is crucial for solving a range of mathematical problems.

Evaluation of Polynomial Functions

Evaluating a polynomial function involves calculating its value at a specific point. This is typically done by substituting the given value of the variable into the polynomial. However, this can become cumbersome with more complex polynomials.

For instance, to evaluate \( f(x) = 4x^2 - 2x + 9 \) at \( x = -3 \) directly, you would substitute and calculate:

• First, calculate \( 4(-3)^2 = 4 \times 9 = 36 \).
• Next, \( -2 \times (-3) = 6 \).
• Finally, add them up: \( 36 + 6 + 9 = 51 \).

Synthetic division streamlines this process, bypassing direct substitution and calculating the remainder instead. This remainder gives us the function's value at the point. This method is particularly helpful for checking work in algebra problems or when dealing with polynomial expressions in calculus.