Problem 34 - Algebra for College Students

Question

Use synthetic division to determine the quotient and remainder for each problem. $$ \left(x^{5}+1\right) \div(x-1) $$

Step-by-Step Solution

Answer

The quotient is \(x^4 + x^3 + x^2 + x + 1\) and the remainder is 2.

Step 1: Set up Synthetic Division

First, we identify the coefficients of the polynomial \(x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1\). We will use synthetic division by the divisor \(x - 1\), so the value we use for division is 1.

Step 2: Write the Coefficients

Write down the coefficients of the dividend polynomial: \([1, 0, 0, 0, 0, 1]\). We will perform synthetic division using these coefficients and the value 1 (from \(x - 1\)).

Step 3: Perform the Synthetic Division

1. Bring down the leading coefficient 1.2. Multiply 1 by the divisor value (1) and add to the next coefficient (0), repeat for each coefficient: - Multiply: \(1 \times 1 = 1\), Add: \(0 + 1 = 1\), write 1. - Multiply: \(1 \times 1 = 1\), Add: \(0 + 1 = 1\), write 1. - Multiply: \(1 \times 1 = 1\), Add: \(0 + 1 = 1\), write 1. - Multiply: \(1 \times 1 = 1\), Add: \(0 + 1 = 1\), write 1. - Multiply: \(1 \times 1 = 1\), Add: \(1 + 1 = 2\), write 2.

Step 4: Read the Quotient and Remainder

The results from synthetic division are the coefficients of the quotient polynomial, starting from one degree less than the original:- Quotient: \(x^4 + x^3 + x^2 + x + 1\)- Remainder: 2

Key Concepts

Polynomial Division

Polynomial division is a method used to divide polynomials, similar to how you divide numbers. There are different techniques under this umbrella, including long division and synthetic division.

This method helps to simplify the process of dividing one polynomial by another, especially when dealing with complex expressions. In our example, we used synthetic division to simplify \(x^5 + 1\) by \(x - 1\).

Here's a simplified approach for polynomial division using synthetic division:

• It requires less writing compared to long division, making it faster and more efficient.
• It works best when dividing by linear terms of the form \(x - c\).
• By focusing only on coefficients, this method reduces potential calculation errors.

Understanding polynomial division allows you to solve complex algebraic expressions more easily, offering practical applications in calculus and beyond.

Remainder Theorem

The Remainder Theorem is a handy tool in polynomial algebra. It states that when a polynomial \(f(x)\) is divided by \(x - c\), the remainder of this division is \(f(c)\).

This theorem provides a quick way to determine the remainder, without performing the full division explicitly:

• It saves time, particularly when dealing with higher-degree polynomials.
• In our original problem, \(f(x) = x^5 + 1\) and \(c = 1\). Substituting \(c\) into \(f(x)\) gives you the remainder:

\[f(1) = 1^5 + 1 = 2\]By applying the Remainder Theorem, you can verify the remainder without completing the synthetic division.

Algebraic Expressions

Algebraic expressions are combinations of variables, coefficients, and constants, formed using mathematical operations like addition, subtraction, multiplication, and division.

Expressions, such as \(x^5 + 1\), are foundational in mathematics and play a crucial role in various fields of study, such as algebra and calculus.

To work effectively with algebraic expressions:

• Understand the role of coefficients and how they change through operations like division.
• Learn how expressions can be manipulated to simplify or solve an equation.
• Master concepts like polynomial division and the remainder theorem to tackle complex problems.

By diving into the world of algebraic expressions, you build a strong mathematical foundation for advanced theoretical and practical applications.