Problem 8 - Algebra for College Students

Question

For Problems 1-10, find \(f(c)\) by (a) evaluating \(f(c)\) directly, and (b) using synthetic division and the remainder theorem. $$ f(n)=8 n^{3}-39 n^{2}-7 n-1 \text { and } c=5 $$

Step-by-Step Solution

Answer

The value of \(f(5)\) is -11.

Step 1: Evaluating Directly

First, we plug in the value of \(c = 5\) directly into the function \(f(n) = 8n^3 - 39n^2 - 7n - 1\). Calculating each part:1. \(8(5)^3 = 8(125) = 1000\)2. \(-39(5)^2 = -39(25) = -975\)3. \(-7(5) = -35\)4. \(-1\)Now add them together:\[1000 + (-975) + (-35) - 1 = -11\]Thus, \(f(5) = -11\).

Step 2: Using Synthetic Division

Now, let's determine \(f(5)\) using synthetic division. Synthetic division can also verify our result by finding the remainder.Set up the synthetic division:- Coefficients of \(f(n)\): \(8, -39, -7, -1\)- Divisor: \(c = 5\)1. Bring down the leading coefficient: 8.2. Multiply 8 by 5, write 40 under the next coefficient: -39. Add: \(-39 + 40 = 1\).3. Multiply 1 by 5, write 5 under the next coefficient: -7. Add: \(-7 + 5 = -2\).4. Multiply -2 by 5, write -10 under the next coefficient: -1. Add: \(-1 + (-10) = -11\).The remainder is \(-11\), which indicates \(f(5) = -11\).

Step 3: Conclusion

Both direct evaluation and synthetic division confirmed that \(f(5) = -11\).

Key Concepts

Synthetic Division

Synthetic Division is a streamlined technique used specifically to divide polynomials when you have a linear divisor of the form \( x - c \). It's much quicker and simpler than traditional long division, particularly when we only care about the remainder. This method is especially useful with polynomial functions to evaluate a function at specific points, like \( c = 5 \) in our example.

To perform synthetic division, you need:

• The coefficients of the polynomial, including zeros for missing terms.
• The zero of the divisor in the form \( x - c \), which is simply \( c \).

Here's a quick breakdown of the steps involved:1. Write down the coefficients of the polynomial.2. Place the value of \( c \) outside a box.3. Bring the first coefficient down.4. Multiply it by \( c \), and add to the next coefficient.5. Repeat the multiply and add process for all coefficients.6. The final number is the remainder, and conveniently also \( f(c) \).

Synthetic division not only provides an efficient way to divide but also helps verify results, as seen in the calculation of \( f(5) \) in our example.

Polynomial Functions

Polynomial functions are expressions that involve sums of powers of variables with coefficients. They are a fundamental concept in algebra and come in various forms, such as linear, quadratic, cubic, and so on. The general form of a polynomial function is: \[ f(x) = a_n x^n + a_{n-1} x^{n-1} + \, \ldots \, + a_1 x + a_0 \]where \( a_n \) are constants known as coefficients, and \( n \) is a non-negative integer called the degree of the polynomial.

In our specific problem, the polynomial \( f(n) = 8 n^3 - 39 n^2 - 7 n - 1 \) is a cubic polynomial, given by its highest exponent of 3.

Cubic polynomials often arise in real-world applications, from physics to engineering, due to their capability to express complex relationships that aren't merely linear or quadratic. Understanding their behavior and roots can be vital in both theoretical and practical contexts.

Function Evaluation

Function evaluation is the process of finding the value of a function for a given input. When you evaluate a function like \( f(c) \), you're substituting a specific value for the variable, then simplifying using arithmetic to find the outcome.

In the example, we had \( f(n) = 8n^3 - 39n^2 - 7n - 1 \), and the task was to find \( f(5) \). This meant substituting 5 for every \( n \) in the polynomial. After substituting, you perform each arithmetic operation:

• Calculate \( 8(5)^3 \)
• Subtract \( 39(5)^2 \)
• Subtract \( 7(5) \)
• Finally, subtract 1

The final result, \( -11 \), represents \( f(5) \).

Evaluating functions is crucial not just for solving polynomial functions but also for understanding how a change in input affects the output. This is central to the study of mathematics and functions' applications across different fields.