Problem 33 - 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 0.

Step 1: Set Up the Division

To perform synthetic division on \(x^5 + 1\) divided by \(x + 1\), we first need to set up our synthetic division using the coefficients of the polynomial \(x^5 + 1\). The coefficients are \(1, 0, 0, 0, 0, 1\) because all terms except for the first and last are missing. The divisor \(x + 1\) implies a root of \(-1\) for synthetic division.

Step 2: Begin Synthetic Division Process

Write down the coefficients \([1, 0, 0, 0, 0, 1]\) horizontally. The value \(-1\) is placed to the left. Bring down the leading coefficient, which is 1, to the bottom row.

Step 3: Multiply and Add

Multiply \(-1\) by the value just written under the line (1). Place the result \(-1\) under the second coefficient (0). Add this to get the new value in the third row, which is \(0 + (-1) = -1\).

Step 4: Repeat Multiply and Add

Continue this process for each coefficient:- Multiply \(-1\) by the new value under each column: -2nd: \(-1 \times -1 = 1\) 3rd: \(-1 \times 1 = -1\) 4th: \(-1 \times -1 = 1\) 5th: \(-1 \times 1 = -1\).- Add each product to the coefficient above it: - 2nd: \(0 + 1 = 1\) 3rd: \(0 - 1 = -1\) 4th: \(0 + 1 = 1\) 5th: \(0 - 1 = -1\) 6th: \(1 - 1 = 0\). The set of results under the line will be \([1, -1, 1, -1, 1, 0]\).

Step 5: Interpret the Results

The bottom row of results from left to right is the quotient and the last value is the remainder. Therefore, the quotient polynomial is \(1x^4 - 1x^3 + 1x^2 - 1x + 1\) and the remainder is \(0\).

Key Concepts

Polynomial Division

Polynomial division is a method used to divide a polynomial by another polynomial of the same or lower degree. In our case, we're dividing \(x^5 + 1\) by \(x + 1\). The process is similar to long division that you might use with numbers, but it involves variables and coefficients instead.

There are two main methods for polynomial division: long division and synthetic division. Long division is more straightforward when polynomials have missing terms or if complexities arise. Synthetic division, however, is simpler and quicker, intended mostly for polynomials with linear divisors of the form \(x - c\).

When performing synthetic division, you only need to use the coefficients of the polynomial you're dividing. This method drastically reduces the complexity of your calculations.

• Synthetic division can only be used for divisors of the form \(x - c\).
• It simplifies the division process by focusing on coefficients.
• It's faster and often more convenient than long division for certain types of problems.

Remainder Theorem

The Remainder Theorem is a useful concept that connects polynomial division with polynomial evaluation. According to this theorem, if you divide a polynomial \(f(x)\) by a linear divisor \(x - c\), then the remainder of this division is \(f(c)\).

In simpler terms, the remainder you get when dividing is exactly what you'd get if you substitute \(c\) into the polynomial function. This is particularly intriguing because it allows you to quickly check the remainder without fully performing division, provided \(c\) is straightforward to substitute.

In our given exercise, after performing synthetic division, the remainder was 0. This indicates that \(-1\) is indeed a root of the polynomial \(x^5 + 1\). Here's a quick checklist to remember about the Remainder Theorem:

• The theorem simplifies finding the remainder of a polynomial division.
• If the remainder is 0, \(c\) is a root of the polynomial.
• This theorem enhances understanding of the relationship between roots and polynomials.

Algebraic Expressions

Algebraic expressions are a fundamental component of algebra, consisting of numbers, variables, and operations. Understanding how to manipulate and simplify these expressions is crucial in solving numerous types of equations, including polynomial ones.

Polynomials are a type of algebraic expression made up of terms that are only whole number powers of \(x\), like \(x^5\), \(x^3\), or \(x^2\). Each term in the polynomial has a coefficient (the number in front of the \(x\) term), which can be seen in our exercise with coefficients like 1 and 0 in \(x^5 + 1\).

Simplifying and reworking these expressions through division or other operations is necessary for finding roots and factoring polynomials. Here are some tips to handle algebraic expressions:

• Always arrange expressions in descending order of the power of variables.
• Combine like terms to simplify expressions.
• Use factoring techniques for breaking down complex expressions.