Problem 42 - Precalculus Mathematics for Calculus

Question

Use synthetic division and the Remainder Theorem to evaluate \(P(c)\). $$P(x)=x^{3}-x^{2}+x+5, \quad c=-1$$

Step-by-Step Solution

Answer

\(P(-1) = 2\).

Step 1: Setup Synthetic Division

First, write down the coefficients of the polynomial \(P(x) = x^3 - x^2 + x + 5\). These coefficients are \([1, -1, 1, 5]\). Next, write the value of \(c\) which is \(-1\) to the left.

Step 2: Perform Synthetic Division

Bring down the first coefficient as is. Multiply this coefficient by \(c\), which is \(-1\), and add to the next coefficient. Continue this process until you reach the last term. Since we are looking for \(P(c)\), the final value after this process is the remainder.

Step 3: Calculate Using Synthetic Division

1. Bring down the \(1\).2. Multiply by \(-1\) and add to \(-1\) yielding \(-2\).3. Multiply \(-2\) by \(-1\) and add to \(1\) yielding \(3\).4. Multiply \(3\) by \(-1\) and add to \(5\) yielding \(2\).So, the remainder or \(P(-1)\) is \(2\).

Step 4: Conclusion from Remainder Theorem

According to the Remainder Theorem, the remainder from the division process is actually \(P(c)\). Thus, \(P(-1) = 2\).

Key Concepts

Remainder Theorem

The Remainder Theorem is a key concept in polynomial arithmetic. It states that when a polynomial \( P(x) \) is divided by a linear divisor \((x - c)\), the remainder of this division process is precisely \( P(c) \). This provides a quick way to evaluate polynomials without substituting and calculating each term manually.
In our example, we have \( P(x) = x^3 - x^2 + x + 5 \) and want to find \( P(-1) \). By using synthetic division as outlined, we see that the remainder from the division process is \( 2 \). So, according to the Remainder Theorem, \( P(-1) = 2 \).

• It connects the concept of evaluating a polynomial at a particular point with the remainder of a division.
• It simplifies calculations significantly when compared to direct substitution.
• It helps in identifying the roots of the polynomial easily when the remainder is zero.

Understanding the Remainder Theorem makes operations with polynomials more efficient and opens the door to other advanced topics in algebra.

Polynomial Division

Polynomial division is analogous to arithmetic division but with polynomials. When dividing one polynomial by another, the result typically consists of a quotient and a remainder. There are different methods to achieve this division, such as long division and synthetic division, with synthetic division being more efficient for dividing by a linear polynomial.
Synthetic division is particularly useful when the divisor is in the form \((x - c)\). It involves using only the coefficients of the polynomial, simplifying the division process while maintaining accuracy. In our problem, we utilize synthetic division due to its advantages in handling linear expressions.

• Synthetic division requires less writing and calculation compared to the traditional long division.
• It is faster and less prone to small calculation errors, making it ideal for evaluating polynomials efficiently.
• Helps in the quick determination of remainders, simplifying the application of the Remainder Theorem.

Polynomial division with synthetic division is a foundational technique that enhances efficiency in solving polynomial equations and is crucial for understanding higher-level algebraic manipulations.

Evaluation of Polynomials

Evaluating a polynomial means calculating the value of the polynomial function for a given input. This is a frequent and crucial task in algebra, as it forms the basis for graphing functions, solving equations, and modeling real-world scenarios.
In the context of synthetic division, evaluation is made easier. By performing synthetic division with \( x = c \), as in our problem where \( P(c) \) was evaluated at \( c = -1 \), the remainder directly provides the evaluation value \( P(-1) = 2 \).

• Direct evaluation through traditional substitution can be time-consuming for higher degree polynomials.
• Synthetic division streamlines this process by providing an organized approach to quickly finding values at specific points.
• Accurate polynomial evaluation is crucial in calculus, applied sciences, and various fields of engineering.

Mastering polynomial evaluation, especially through effective techniques, is vital for students to progress in mathematics, as it lays the groundwork for exploring complex functions and algebraic structures.