Problem 21 - Basic Technical Mathematics with Calculus

Question

Use the factor theorem to determine whether or not the second expression is a factor of the first expression. Do not use synthetic division. $$x^{51}-2 x-1, x+1$$

Step-by-Step Solution

Answer

Yes, \( x+1 \) is a factor of \( x^{51}-2x-1 \).

Step 1: Recall the Factor Theorem

The Factor Theorem states that a polynomial \( f(x) \) has a factor \( (x - c) \) if and only if \( f(c) = 0 \). This means, to check if \( x+1 \) is a factor, we need to substitute \( x = -1 \) into the polynomial \( f(x) = x^{51} - 2x - 1 \) and see if it equals zero.

Step 2: Substitute \( x = -1 \) into \( f(x) \)

We substitute \( x = -1 \) into the polynomial: \( f(-1) = (-1)^{51} - 2(-1) - 1 \).

Step 3: Simplify the Expression

Calculate \( (-1)^{51} \). Since 51 is an odd number, \((-1)^{51} = -1\). Then calculate \(-2(-1) = 2\). Thus, the expression becomes \(-1 + 2 - 1\).

Step 4: Calculate the Result

Simplify \(-1 + 2 - 1 = 0\).

Step 5: Conclusion based on the Factor Theorem

Since \( f(-1) = 0 \), the expression \( x+1 \) is a factor of \( x^{51}-2x-1 \) according to the Factor Theorem.

Key Concepts

Polynomial Division

Polynomial division is similar to long division but works with polynomials instead of numbers. It is a method to divide one polynomial by another, attaining a quotient and possibly a remainder. In our exercise, we are checking if a polynomial can be divided evenly by another polynomial, which, if true, implies that the divisor is a factor.
To perform polynomial division, arrange terms in descending order of their exponents. Align the polynomial to its respective powers of the variable. Each step of the division involves dividing the leading term of the dividend by the leading term of the divisor. This gives you part of the quotient.

• Write the quotient above the line.
• Multiply the entire divisor by the current quotient term.
• Subtract the result from the dividend.

Repeat these steps until you've processed every term or as far as you can go. If there’s no remainder, it implies that the divisor is a factor of the dividend.

Substitution in Polynomials

Substitution in polynomials involves replacing the variable with a numeric value. This is crucial when using the Factor Theorem, which is our main strategy here. By substituting a specific value, we can verify if our substitution results in zero, indicating that the substituted value corresponds to a factorization point.
In our specific problem, we substituted the value \(x = -1\). This arises from rewriting \(x + 1\) as \(x - (-1)\), aligning with the Factor Theorem's requirement to test \(f(c) = 0\) for the expression to be a factor. Follow these steps:

• Replace every instance of the variable in the polynomial with the chosen number.
• Calculate each term's result step by step.
• Simplify the final expression to find the result.

If the result is zero, the polynomial exactly divides, confirming the factor.

Odd and Even Exponents

Understanding the behavior of odd and even exponents is key to managing polynomial expressions effectively, especially during substitutions. When evaluating polynomial terms with exponents:- **Odd Exponents**: If a base is negative, raising it to an odd exponent results in a negative value. For example, \((-1)^{51} = -1\) because raising \(-1\) to any odd power keeps the sign negative.- **Even Exponents**: In contrast, a negative base raised to an even exponent results in a positive value. For instance, \((-1)^2 = 1\), as the negative signs cancel out each other.In the problem, recognizing \(51\) as an odd number helped determine that \((-1)^{51} = -1\). Each exponent's property plays a role in determining whether the entire polynomial equates to zero after substitution, thereby aiding in confirming or denying factorization.