Problem 9 - 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. $$ f(n)=2 n^{5}-1 \text { and } c=-2 $$

Step-by-Step Solution

Answer

\(f(-2) = -65\).

Step 1: Evaluate f(c) Directly

First, evaluate the function \(f(n) = 2n^5 - 1\) directly by substituting \(n = c = -2\):\[f(-2) = 2(-2)^5 - 1\]Calculate \((-2)^5\):\[(-2)^5 = -32\]Now substitute back to find:\[f(-2) = 2(-32) - 1 = -64 - 1 = -65\]Therefore, by directly evaluating \(f(c)\), we find \(f(-2) = -65\).

Step 2: Set Up Synthetic Division

To use synthetic division, write down the coefficients of \(f(n) = 2n^5 - 1\). Notice all missing degree terms have coefficients of 0. So, the polynomial \(f(n)\) has coefficients:\[[2, 0, 0, 0, 0, -1]\]Use \(c = -2\) as the divisor.

Step 3: Perform Synthetic Division

Start synthetic division by writing \(-2\) on the left, and perform the operations:1. Bring down the 2: \[ \underline{2}\]2. Multiply \(-2\) by 2, and add the result to the next coefficient 0: \[ 2 \times (-2) = -4\]3. Write \(-4\) under the next column, and add to 0: \[ 0 + (-4) = -4\]4. Multiply \(-2\) by \(-4\), and add to the next 0: \[ -4 \times (-2) = 8\]5. Write 8 under the next column, and add to 0: \[ 0 + 8 = 8\]6. Repeat the process for the remaining terms until completion: \[ -2 \times 8 = -16 \ 0 + (-16) = -16 \ -2 \times (-16) = 32 \ 0 + 32 = 32 \ -2 \times 32 = -64 \ (-1) + (-64) = -65 \]

Step 4: Interpret the Remainder

In synthetic division, the last number is the remainder. From the division steps, we have found the last term to be \(-65\), thus confirming the result:By synthetic division, the remainder is \(f(-2) = -65\). From the Remainder Theorem, this is \(f(c)\).

Key Concepts

Remainder Theorem

The Remainder Theorem is a powerful tool in polynomial algebra. It provides a simple way to evaluate a polynomial function at a given point. When you divide a polynomial \( f(n) \) by \( (n - c) \) using synthetic division, the remainder you get is the value of the polynomial at \( n = c \). This means that if you perform the division and the remainder is \(-65\), it confirms that \( f(c) \) or \( f(-2) \) in our example is indeed \(-65\).

This theorem can save time, especially with complex polynomials, by avoiding direct substitution and multiple calculations. It's also a nifty verification method to ensure that your calculations from synthetic division are correct.

Polynomial Evaluation

Evaluating a polynomial means finding its value for a specific input, which is often represented as \( f(c) \). This requires substituting the given value of \( c \) into the polynomial equation and calculating the result.

For the polynomial \( f(n) = 2n^5 - 1 \) and \( c = -2 \), we substitute \(-2\) into the equation:
\[\begin{align*} f(-2) & = 2(-2)^5 - 1 \ & = 2(-32) - 1 \ & = -64 - 1 \ & = -65. \end{align*}\] By doing this, we've directly evaluated \( f(c) \) to find that it gives a value of \(-65\). This method involves straightforward computation but can be prone to manual calculation errors, especially with high-degree terms.

Direct Substitution

Direct substitution is often the most straightforward way to evaluate a polynomial. You simply replace every occurrence of the variable \( n \) in the polynomial with the given value \( c \).

In our example, we took the polynomial \( f(n) = 2n^5 - 1 \) and directly substituted \(-2\) for \( n \) to get \( f(-2) = 2(-2)^5 - 1 \). By calculating \((-2)^5 = -32\) and then multiplying by 2, you end up with \(-64\). Subtracting 1 finally gives \(-65\).

This method is direct but requires careful calculation of powers and arithmetic. For complex polynomials or higher powers, it can be cumbersome, and errors might sneak in, but it's a great place to start for understanding how changes in \( n \) affect the polynomial's output.