Problem 11 - Algebra 2 and Trigonometry

Question

In \(3-18,\) find all roots of each given function by factoring or by using the quadratic formula. $$ f(x)=x^{4}-81 $$

Step-by-Step Solution

Answer

Roots are \(x = 3, -3, 3i, -3i\).

Step 1: Identify the type of expression

We start with the expression \(f(x) = x^4 - 81\). This is a difference of squares because \(81\) can be represented as \(9^2\).

Step 2: Express as a difference of squares

Recognize that \(x^4 - 81\) can be rewritten as \((x^2)^2 - 9^2\). This is a classic difference of squares, which can be factored using the formula \(a^2 - b^2 = (a - b)(a + b)\).

Step 3: Factor the difference of squares

Using the difference of squares formula, factor \((x^2)^2 - 9^2\) as \((x^2 - 9)(x^2 + 9)\).

Step 4: Factor further if possible

The expression \(x^2 - 9\) can be further factored, since \(-9 = -3^2\), resulting in \((x - 3)(x + 3)\). The expression \(x^2 + 9\) cannot be factored using real numbers since it represents a sum of squares.

Step 5: Solve for roots

Set each factor equal to zero and solve: For \(x^2 - 9 = 0\), we get \(x = 3\) and \(x = -3\). For \(x^2 + 9 = 0\), solve it using imaginary numbers: \(x^2 = -9\), hence \(x = \pm 3i\).

Step 6: Compile the roots

The complete set of roots of the equation \(f(x) = x^4 - 81\) are \(x = 3, -3, 3i, -3i\).

Key Concepts

Difference of Squares

The concept of the difference of squares is essential in algebra and occurs when you have an expression of the form \(a^2 - b^2\). Breaking it down, it consists of two perfect squares separated by a subtraction sign. This pattern is quite versatile and can be factored directly using the formula:

• \(a^2 - b^2 = (a - b)(a + b)\)

This simplification method is powerful in simplifying and solving polynomial equations by reducing higher powers into manageable factors. For instance, in the exercise, \(x^4 - 81\) was identified as a difference of squares because \(x^4\) is \((x^2)^2\) and \(81\) is \(9^2\). This allowed us to transform it into \((x^2 - 9)(x^2 + 9)\).

Using this technique often simplifies what might initially appear as more complicated problems, making it easier to find solutions or further factor the expression.

Factoring Expressions

Factoring expressions is a critical technique used to simplify polynomials and solve equations. The goal is to break down a complex expression into products of simpler ones. Particularly, factoring is beneficial in solving quadratic equations or higher degree polynomials.

• The first step is to identify if the expression fits a recognizable pattern, such as the difference of squares or perfect square trinomials.
• Then, apply the appropriate factoring formula or method.
• For example, \((x^2 - 9)\) from our exercise was factored into \((x - 3)(x + 3)\), which involved recognizing that \(9\) is a perfect square (\(3^2\)).

Factoring transforms complex expressions into simpler factors that can easily yield solutions when set to zero. Recognizing patterns like these and using them appropriately is key to mastering algebraic manipulations.

Quadratic Formula

The quadratic formula is an invaluable tool in solving quadratic equations of the form \(ax^2 + bx + c = 0\). This formula provides a straightforward solution, particularly when other factoring methods become cumbersome or impossible. The standard quadratic formula is:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In cases where factoring is not apparent or when dealing with irrational or complex roots, the quadratic formula acts as a universal method for finding roots. Although not directly applied in every factoring task, knowing this formula backs up your toolkit with powerful problem-solving capacity. It's crucial when dealing with expressions like \(x^2 + 9\), which lead to complex numbers due to the negative discriminant.

Imaginary Numbers

Imaginary numbers arise when dealing with the square roots of negative numbers, which are not defined in the realm of real numbers. The foundation of imaginary numbers is built on \(i\), representing \(\sqrt{-1}\). This invention allows for solutions to equations that otherwise have no real number solutions.

• In our exercise, solving \(x^2 + 9 = 0\) presented a need for imaginary numbers because \(x^2 = -9\) translates to \(x = \pm \sqrt{-9}\).
• The result is \(x = \pm 3i\), where \(3i\) and \(-3i\) are imaginary numbers.

Understanding imaginary numbers is crucial when working with polynomials that result in non-real roots. They extend the capabilities of algebra and are pivotal in advanced mathematics, such as complex number theory.