Problem 93 - Algebra and Trigonometry

Question

Find all zeros of \(f(x)=2 x^{3}+x^{2}-13 x+6 .\) (Section 3.4 Example \(3)\)

Step-by-Step Solution

Answer

The zeros of the function \(f(x) = 2x^3 + x^2 - 13x + 6\) are \(x = 1\), \(x = -2\), and \(x = 1.5\).

Step 1: Find an initial root

The first step is to find a possible root. We can do this by checking a few values of \(x\). By trial and error, it is found that \(x = 1\) is a root as it makes \(f(x) = 0\). This is our initial root.

Step 2: Perform Polynomial Division

Since \(x = 1\) is a root, \((x - 1)\) is a factor. So, we divide the polynomial \(f(x) = 2x^3 + x^2 - 13x + 6\) by \((x - 1)\) to find the other factors. The result is \(f(x) = (x - 1)(2x^2 + 3x - 6)\).

Step 3: Solve the quadratic equation

Now we can solve the quadratic equation \(2x^2 + 3x - 6 = 0\) for \(x\). To solve for the roots of this quadratic equation, we use the quadratic formula: \(x = [-b ± sqrt(b^2 - 4ac)] / 2a\). Substituting \(a = 2\), \(b = 3\), and \(c = -6\) into the formula, we get two more roots of the polynomial, which are \(x = -2\) and \(x = 1.5\).

Step 4: Combine all roots

Combining all our roots, the zeros of \(f(x)\) are \(x = 1\), \(x = -2\), and \(x = 1.5\).

Key Concepts

Polynomial Division

Polynomial division is a crucial step when finding zeros of polynomial equations, especially when you identify a single root and need to uncover the others. Imagine polynomial division as comparable to long division we use in arithmetic, but instead, it's applied to dividing a polynomial by another polynomial, often of lower degree.

When you find a root, like in our example where we found that \(x = 1\) is a root of the polynomial \(f(x) = 2x^3 + x^2 - 13x + 6\), this tells us that \((x - 1)\) is a factor of \(f(x)\). By dividing \(f(x)\) by \((x - 1)\), this breaks down the polynomial into simpler parts, revealing other potential factors, something like peeling back layers in an onion.

• Start by writing down the terms of the polynomial in decreasing order of their powers.
• Divide the first term of the polynomial by the first term of the divisor.
• Multiply the entire divisor by the result you obtained and subtract this product from the polynomial.
• Repeat the process with the resulting polynomial until the degree of the remainder is less than the degree of the divisor.

In our task, dividing \(f(x)\) by \((x - 1)\) simplifies it to \((x - 1)(2x^2 + 3x - 6)\), making the next steps to find zeros much more manageable.

Quadratic Formula

The quadratic formula is a universal tool for finding roots of any quadratic equation of the form \(ax^2 + bx + c = 0\). Its beauty lies in its ability to solve quadratic equations that are otherwise challenging to factor.

We use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation. It saves time and reduces errors compared to trial and error or complex factoring.

• First, ensure your equation is in the standard form \(ax^2 + bx + c = 0\).
• Identify the coefficients: \(a\) is the coefficient of \(x^2\), \(b\) is for \(x\), and \(c\) is the constant.
• Substitute these values into the quadratic formula.
• Simplify under the square root (discriminant) and ensure you consider both the `+` and `-` options for potential solutions.

For our example polynomial, we reach the equation \(2x^2 + 3x - 6 = 0\), where \(a = 2\), \(b = 3\), and \(c = -6\). Substituting these into the quadratic formula gives us the roots \(x = -2\) and \(x = 1.5\), solving the quadratic part of the polynomial equation.

Roots of Polynomial Equations

Finding the roots of polynomial equations means solving for the values of \(x\) where the polynomial equals zero. These solutions or roots can be real or complex numbers.

For any polynomial, the Fundamental Theorem of Algebra tells us that a polynomial of degree \(n\) will have exactly \(n\) roots, though some may be repeated (termed "multiplicity").

• Start by identifying any obvious roots through simple substitution or inspection. Integers or factors of the constant term are good candidates.
• Use methods such as factoring, polynomial division, and synthetic division to break down the polynomial.
• For quadratic sections, apply the quadratic formula, or completing the square can also be effective.
• Check each solution by plugging it back into the original equation.

In our exercise lesson, we found the roots for the cubic polynomial \(2x^3 + x^2 - 13x + 6 = 0\) by first identifying \(x = 1\) through trial and error. After performing polynomial division and using the quadratic formula, we discovered the complete set of zeros: \(x = 1\), \(x = -2\), and \(x = 1.5\). Mastering these processes not only helps you solve the current polynomial but equips you with a reliable strategy for tackling future polynomial equations.