Problem 17 - Basic Technical Mathematics with Calculus

Question

Solve the given equations using synthetic division, given the roots indicated. $$6 x^{4}+5 x^{3}-15 x^{2}+4=0 \quad\left(r_{1}=-\frac{1}{2}, r_{2}=\frac{2}{3}\right)$$

Step-by-Step Solution

Answer

The polynomial has roots \(-\frac{1}{2}, \frac{2}{3}, 1, -2\).

Step 1: Setup for Synthetic Division with the first root

To use synthetic division with the root \( r_1 = -\frac{1}{2} \), let's write down the coefficients of the polynomial: 6 (for \( x^4 \)), 5 (for \( x^3 \)), -15 (for \( x^2 \)), and 4 (constant term). These coefficients form the row for the synthetic division.

Step 2: Perform Synthetic Division with \( r_1 = -\frac{1}{2} \)

Start the synthetic division by bringing down the leading coefficient (6) in front of the bar. Multiply it by \( -\frac{1}{2} \) and add to the second coefficient \( 5 \), repeating the process until the end of the row. You should get a remainder of zero if \( -\frac{1}{2} \) is a root of the polynomial. The intermediate results will be: \( 6, 2, -16, 12, 0 \). The zero at the end confirms \( -\frac{1}{2} \) is a root.

Step 3: Setup for Synthetic Division with the second root

Use the coefficients obtained from the previous step \( 6, 2, -16, 12 \), excluding the remainder, for the next synthetic division step. Now use \( r_2 = \frac{2}{3} \) to perform synthetic division.

Step 4: Perform Synthetic Division with \( r_2 = \frac{2}{3} \)

As before, bring down the initial coefficient (6). Multiply by \( \frac{2}{3} \) and add to the next coefficient. Continue this process through the entire dataset. The process results in coefficients \( 6, 6, -12, 0 \), confirming zero remainder and verifying \( \frac{2}{3} \) is a root.

Step 5: Find Remaining Quadratic Equation

Now that we have performed synthetic division twice, we find the reduced polynomial from the division is \( 6x^2 + 6x - 12 \). It can potentially have more roots.

Step 6: Simplify and Solve the Quadratic Equation

Factor the quadratic equation \( 6x^2 + 6x - 12 \). Simplifying, this becomes \( x^2 + x - 2 = 0 \). Factor to \( (x - 1)(x + 2) = 0 \). The solutions to these factors are \( x = 1, -2 \).

Key Concepts

Polynomial Roots

Understanding polynomial roots is essential for mastering algebra. These roots, sometimes called zeros, are the values for which a given polynomial equals zero. In other words, if you substitute the roots in the place of the variable in the polynomial, the result should be zero.

The polynomial provided is a fourth-degree polynomial, meaning it can have up to four roots. If given roots such as \(-\frac{1}{2}\) and \(\frac{2}{3}\), these values satisfy the polynomial equation, making the overall expression zero.

• Identifying the roots simplifies complex polynomial equations into solvable units.
• Synthetic division aids in verifying whether a given value is indeed a root.

With the roots \(-\frac{1}{2}\) and \(\frac{2}{3}\), synthetic division aids in reducing the polynomial into a simpler form by removing these roots.

Quadratic Equation

Quadratic equations are polynomials of degree two, typically expressed in the form \(ax^2 + bx + c = 0\). Their simplicity makes them a prime candidate for factoring or other solving techniques.

Through synthetic division, the original polynomial is reduced to a quadratic equation \(6x^2 + 6x - 12\). Solving this involves a few steps such as simplifying and, ideally, factoring it. The quadratic still holds potential roots that complete the four solutions anticipated for a quartic equation.

In our exercise, the quadratic simplifies to \(x^2 + x - 2 = 0\). Recognizing this step is crucial as it transitions from a higher degree equation to quadratic, suited for familiar solving methods.

Factoring

Factoring is a critical step in finding the solution to quadratic equations. When you factor, you essentially break down the equation into the product of simpler equations. For example, to factor \(x^2 + x - 2 = 0\), you look for numbers that multiply to the constant term, \(-2\), and add to the linear coefficient, \(1\).

Factoring leads to:

• \((x - 1)(x + 2) = 0\)

By factoring in this manner, identifying further solutions becomes more visible.

• These simplified products translate into the roots \(x = 1\) and \(x = -2\).

This method underlines the power of factoring in swiftly determining all potential solutions of quadratic equations derived during polynomial exercises.