Problem 30 - Algebra for College Students

Question

For Problems \(21-34\), use the factor theorem to help answer some questions about factors. $$ \text { Is } x+4 \text { a factor of } 2 x^{3}+9 x^{2}-5 x-39 ? $$

Step-by-Step Solution

Answer

No, \(x+4\) is not a factor of the polynomial.

Step 1: Understand the Problem

We need to determine if \(x+4\) is a factor of the polynomial \(2x^3 + 9x^2 - 5x - 39\). According to the Factor Theorem, \(x + 4\) is a factor of the polynomial if \(p(-4) = 0\), where \(p(x) = 2x^3 + 9x^2 - 5x - 39\).

Step 2: Substitute -4 into the Polynomial

Calculate \(p(-4)\) by substituting \(-4\) into the polynomial: \[ p(-4) = 2(-4)^3 + 9(-4)^2 - 5(-4) - 39 \]

Step 3: Calculate Each Term

First, calculate each term separately: - \((-4)^3 = -64\)- \((-4)^2 = 16\)Now substitute: - \(2(-64) = -128\)- \(9(16) = 144\)- \(-5(-4) = 20\)- \(-39\) remains the same.

Step 4: Simplify and Sum the Terms

Combine all the terms to find \(p(-4)\):\[ p(-4) = -128 + 144 + 20 - 39 \] Perform the addition and subtraction: - \(-128 + 144 = 16\) - \(16 + 20 = 36\) - \(36 - 39 = -3\)

Step 5: Check the Result

Since \(p(-4) = -3\) and not 0, \(x+4\) is not a factor of \(2x^3 + 9x^2 - 5x - 39\) according to the Factor Theorem.

Key Concepts

Polynomial Division

Polynomial division is a useful mathematical technique to simplify complex polynomial expressions. It's similar to long division with numbers, but it involves variables and coefficients. This process helps us find out if one polynomial is a factor of another. Here's how it works in a nutshell:

• We divide the given polynomial by another polynomial, typically a simpler one.
• The goal is to see if there's a remainder after the division.

If the division results in zero remainder, it indicates that the divisor is a factor of the dividend.
This process is particularly handy in algebra to break down complex expressions into simpler factors, making them easier to work with.

Roots of Polynomials

The roots of a polynomial, sometimes known as zeros, are the values of the variable that make the polynomial equal to zero. Essentially, these are the x-values where the polynomial graph intersects the x-axis.

• For a linear polynomial like \(x + 4\), the root is simply the value that makes this expression zero, which is \(-4\) in this case.
• For more complex polynomials, finding roots can involve factoring, using the quadratic formula, or applying numerical methods.

Understanding the roots of a polynomial is critical because it tells us about the behavior of the polynomial function and its potential intercepts on a graph. When solving polynomial equations, determining the roots is often the main goal.

Synthetic Substitution

Synthetic substitution is a quick and efficient method for evaluating a polynomial at a given value. This technique is especially useful when applying the Factor Theorem, as it helps determine if a specific x-value is a root of the polynomial.
Here is how synthetic substitution works in practice:

• Set up a simplified division-like format by listing the coefficients of the polynomial.
• Use the x-value by which you want to evaluate the polynomial (often the potential root) and perform a series of simple calculations.

This method reduces the need for complex calculations and makes it easier to test different values. If the result of these calculations is zero, the x-value is a root of the polynomial, indicating that the divisor associated with this root is a factor of the original polynomial.