Problem 22 - 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. $$x^{7}-128^{-1}, x+2^{-1}$$

Step-by-Step Solution

Answer

No, \( x + 2^{-1} \) is not a factor of \( x^7 - 128^{-1} \).

Step 1: Identify the expressions

We are given the polynomial expression \( f(x) = x^7 - 128^{-1} \) and need to check if \( x + 2^{-1} \) is a factor.

Step 2: Rearrange the factor

According to the factor theorem, \( x + a \) is a factor of \( f(x) \) if \( f(-a) = 0 \). In our case, rearrange \( x + 2^{-1} \) as \( x - (-2^{-1}) \), so \( a = -2^{-1} \).

Step 3: Substitute into the polynomial

To apply the factor theorem, substitute \( -2^{-1} \) into the polynomial: \( f(-2^{-1}) = ((-2^{-1})^7 - 128^{-1}) \).

Step 4: Calculate powers and simplifications

Calculate \((-2^{-1})^7 = -2^{-7}\) and \(128^{-1} = (2^7)^{-1} = 2^{-7}\). Thus, simplify \( f(-2^{-1}) = -2^{-7} - 2^{-7}\).

Step 5: Check the result

Combine the terms: \(-2^{-7} - 2^{-7} = -2(2^{-7}) = -2^{-6}\), which is not zero.

Key Concepts

Polynomial Expression

Polynomial expressions are mathematical expressions consisting of variables raised to whole number powers and coefficients. They can have one or multiple terms, and their general form is given by:\[ f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \]where \( a_n, a_{n-1}, \ldots, a_0 \) are coefficients, and \( n \) is a non-negative integer indicating the degree of the polynomial.In the given exercise, we have the polynomial \( f(x) = x^7 - 128^{-1} \). This expression consists of a variable term \( x^7 \) and a constant term \( -128^{-1} \). Here, the degree of the polynomial is 7, as it is the highest power of \( x \) in the expression.It's important to recognize that polynomials can encompass various forms, and understanding their structure helps in operations like factorization and applying the factor theorem.

Factorization

Factorization is the process of breaking down a complex expression into simpler components or 'factors' that, when multiplied together, give back the original expression. For polynomial expressions, factorization often involves finding expressions that divide the polynomial without leaving a remainder.In this context, the factor theorem plays a pivotal role. It states that if \( f(a) = 0 \) for a polynomial \( f(x) \), then \( x - a \) is a factor of \( f(x) \). Thus, if a polynomial expression can be written as the product of lower-degree polynomials, it can be termed as factorized.For example, determining if \( x + \frac{1}{2} \) is a factor of \( x^7 - 128^{-1} \) involves substituting \( -\frac{1}{2} \) for \( x \) in the polynomial expression and checking if the result is zero. This approach simplifies the process by sidestepping the need for synthetic or long division techniques.

Mathematical Proof

Mathematical proof is a logical argument that verifies the truth of a statement or theorem. In algebra, proving whether an expression is a factor of a polynomial relies heavily on logical reasoning and mathematical concepts, like the factor theorem.To prove or disprove that \( x + \frac{1}{2} \) is a factor of \( x^7 - 128^{-1} \), we must demonstrate that substituting \( -\frac{1}{2} \) into the polynomial results in zero. Here is the process:

• Substitute \( x = -\frac{1}{2} \) into \( f(x) = x^7 - 128^{-1} \) yielding \( f(-\frac{1}{2}) \).
• Calculate the powers: \((-\frac{1}{2})^7 = -\frac{1}{128}\).
• Simplify: \( f(-\frac{1}{2}) = -\frac{1}{128} - \frac{1}{128} = -\frac{2}{128} = -\frac{1}{64} \).

Since the result is not zero, \( x + \frac{1}{2} \) is not a factor of the original polynomial. This deductive process involves applying concepts methodically to reach a conclusion, illustrating the power of mathematical proof in understanding polynomial relationships.