Problem 38 - Basic Technical Mathematics with Calculus

Question

Use the factor theorem and synthetic division to determine whether or not the second expression is a factor of the first. $$6 x^{4}+5 x^{3}-x^{2}+6 x-2 ; \quad 3 x-1$$

Step-by-Step Solution

Answer

The expression \(3x-1\) is a factor of \(6x^{4}+5x^{3}-x^{2}+6x-2\).

Step 1: Verify Factor Form

According to the factor theorem, if \(3x - 1\) is a factor of \(6x^{4} + 5x^{3} - x^{2} + 6x - 2\), then the root of the equation, \(x = \frac{1}{3}\), should set the polynomial equal to zero. We substitute \(x = \frac{1}{3}\) into the polynomial to verify this.

Step 2: Substitute and Simplify

Substitute \(x = \frac{1}{3}\) into \(6x^{4} + 5x^{3} - x^{2} + 6x - 2\):\[6 \left(\frac{1}{3}\right)^4 + 5 \left(\frac{1}{3}\right)^3 - \left(\frac{1}{3}\right)^2 + 6 \left(\frac{1}{3}\right) - 2\]Evaluate each term:\[6 \times \frac{1}{81} + 5 \times \frac{1}{27} - \frac{1}{9} + 2 - 2\]This simplifies to:\[\frac{6}{81} + \frac{15}{81} - \frac{9}{81} + \frac{162}{81} - \frac{162}{81}= \frac{0}{81} = 0\]Since this results in zero, \(3x - 1\) is a factor of the polynomial.

Step 3: Set Up Synthetic Division

For synthetic division, use the root \(x = \frac{1}{3}\). Write the coefficients of the polynomial: \[6, 5, -1, 6, -2\]. Place \(\frac{1}{3}\) outside the synthetic division box.

Step 4: Perform Synthetic Division

Carry down the leading coefficient, \(6\):\[\begin{array}{c|ccccc}\frac{1}{3} & 6 & 5 & -1 & 6 & -2 \ & \downarrow & \ \hline& 6 & 7 & 1 & 7 & 0 \\end{array}\]Follow these steps:1. Multiply \(6\) by \(\frac{1}{3}\) and write the result under the next coefficient: \(2\).2. Add this to the next coefficient \(5\) to get \(7\).3. Repeat for the next position to maintain the dividend construction until completing all coefficients.

Step 5: Check the Remainder

The final result from synthetic division ends with a remainder of 0, confirming that \(3x - 1\) is indeed a factor, as a remainder of zero confirms the factor theorem.

Key Concepts

Synthetic Division

Synthetic Division is a streamlined method of dividing a polynomial by a binomial of the form \(x - c\), where \(c\) is a constant. This method is often preferred over long division because it is concise and less prone to error.
Instead of working with the entire polynomial at each step, synthetic division focuses only on the coefficients, making the process much more efficient. The steps are as follows:

• Identify the root of the divisor, \(c\). For a binomial like \(3x - 1\), it is \(x = \frac{1}{3}\). You use this root in the synthetic division table.
• Write down all the coefficients of the polynomial in descending order of powers, including zeros for any missing terms. For the polynomial \(6x^{4} + 5x^{3} - x^{2} + 6x - 2\), these coefficients are \([6, 5, -1, 6, -2]\).
• Begin the synthetic division by bringing the first coefficient straight down.
• Multiply the root by the number directly above it and add the result to the next coefficient. Continue this step across all coefficients.
• The last number in this process is the remainder. A zero remainder, as in this exercise, confirms that the binomial is a factor of the polynomial.

This method not only helps determine factors efficiently but also simplifies the overall division process, saving time and effort in polynomial computations.

Polynomial Division

Polynomial Division involves dividing one polynomial by another, just as you might divide numbers. There are generally two methods: long division and synthetic division. Understanding these processes will strengthen your mathematical problem-solving skills.
Long Division is similar to the division process for numbers but can become cumbersome with many steps. It requires dividing, multiplying, subtracting, and finally, bringing down the next term from the dividend repeatedly until you've processed all terms.
Synthetic Division streamlines this by focusing on coefficients, as previously explained. Synthetic division, as used in this exercise, makes the division faster, but can only be used in specific scenarios:

• When dividing by a linear binomial of the form \(x - c\).
• When you need a quick way to check if a binomial is a factor.

Both methods can be used effectively to solve polynomial division problems. However, synthetic division is usually preferred for its quickness and simplicity, particularly when dealing with large polynomials or when verifying factors.

Roots of Polynomials

Roots of a polynomial are the solutions for which the polynomial equals zero. They are integral to understanding the behavior and characteristics of the polynomial function.
In the given exercise, the factor theorem is employed. This theorem states that for a polynomial \(P(x)\), \(x = c\) is a root of \(P(x)\) if and only if \((x - c)\) is a factor of the polynomial. In this context, finding a root is synonymous with identifying a factor.
To find the root, substitute different values to see if the resulting equation equals zero. In this exercise, substituting \(x = \frac{1}{3}\) into the polynomial returns zero, confirming it as a root. This root corresponds to the factor \((3x - 1)\).
Understanding the roots of a polynomial is key to solving many algebraic problems.

• They are useful in graphing because they indicate where the polynomial will cross the x-axis.
• They provide insights into the nature of polynomial behavior.
• They help determine whether a polynomial can be factored further.

Overall, roots and factors reveal essential details about the polynomial's structure and solutions.