Problem 23 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find all the zeros of the function and write the polynomial as a product of linear factors. Use a graphing utility to verify your results graphically. (If possible, use the graphing utility to verify the imaginary zeros.) $$f(x)=16 x^{4}-81$$

Step-by-Step Solution

Answer

The zeros of the function \(f(x)=16x^{4}-81\) are \(x=3/2, -3/2\) and \(x=±\frac{3i\sqrt{2}}{4}\). The polynomial factors to \(f(x)=(2x-3)(2x+3)(4x^2+9)\)

Step 1: Factoring the Polynomial

To find the zeroes of the polynomial, it needs to be factored to the simplest form. The function \(f(x)=16x^{4}-81\) can be written as difference of two squares i.e., \(f(x)=(4x^2)^2-(9)^2\). We know that a^2 - b^2 is equal to \((a-b)(a+b)\), so this can be factored as \(f(x)=(4x^2-9)(4x^2+9)\).

Step 2: Factoring the Polynomial Further

The first factor derived in Step 1, i.e., \((4x^2-9)\), can be factored further as it also represents a difference of squares. Factoring it further gives, (\(2x-3\))(\(2x+3\)). Therefore, the complete factoring of \(f(x)\) is now \((2x-3)(2x+3)(4x^2+9)\).

Step 3: Finding the Zeros of the Function

Set the factors derived in Step 2 equal to zero and solve for x. When \((2x-3)=0\), \(x=3/2\). When \((2x+3)=0\), \(x=-3/2\). The zeroes for \((4x^2+9)\) are complex, which can be found by setting \(4x^2=-9\) and solving to give \(x=±\frac{3i}{2\sqrt{2}}\) or \(x=±\frac{3i\sqrt{2}}{4}\).

Key Concepts

Difference of Squares

The concept of the "difference of squares" is a very useful tool in algebra, especially when factoring polynomials. When you have an expression in the form of \(a^2 - b^2\), it can be factored into

• (a - b)(a + b)

This identity helps break down expressions into simpler components. In the provided exercise, the polynomial \(f(x) = 16x^4 - 81\) is an example of a difference of squares.
Here, \(16x^4\) is \((4x^2)^2\) and \(81\) is \(9^2\).
Recognizing this form allows us to rewrite the polynomial as \((4x^2 - 9)(4x^2 + 9)\).
By doing so, it becomes easier to solve complex problems and find the roots of the polynomial.

Complex Zeros

When dealing with polynomial equations, we sometimes encounter "complex zeros". These are roots that involve imaginary numbers. In mathematical terms, an imaginary number is a multiple of \(i\), where \(i\) is defined as the square root of -1. In the step-by-step solution for our exercise, after factorizing the polynomial as

• \((2x-3)(2x+3)(4x^2+9)\)

we notice the polynomial \(4x^2 + 9\) does not have real solutions. This is because solving \(4x^2 = -9\) involves taking the square root of a negative number.
To solve this, divide both sides by 4, yielding \(x^2 = -\frac{9}{4}\).
Taking the square root gives \(x \pm \frac{3i}{2}\). In the end, the complex zeros of this polynomial are \(x = \pm \frac{3i}{2}\). Understanding complex zeros is important because not all polynomial equations have real-number solutions.

Factorization

Factorization is the process of breaking down a more complex expression into simpler factors. In the context of a polynomial, it involves expressing it as a product of simpler polynomials. The main goal is to make it easier to solve the equation, find zeros, or simplify the expression. Starting with the polynomial \(f(x) = 16x^4 - 81\) from the exercise, we use the difference of squares to factorize it into

• \((4x^2 - 9)(4x^2 + 9)\)

However, the process doesn't stop there. We can further break it down. The expression \(4x^2 - 9\) is also a difference of squares.
This means it can be further factorized into

• \((2x - 3)(2x + 3)\)

Through factorization, the original polynomial is simplified, making it easier to identify and solve for the zeros of the polynomial.

Linear Factors

Breaking down polynomials into "linear factors" is a crucial method for easily finding their roots. Linear factors are polynomials of degree one, which typically appear in the form of \(ax + b\). From the exercise, by factorizing and using the difference of squares, we decompose \(16x^4 - 81\) ultimately into its linear components as:

• \((2x - 3)(2x + 3)\)

In this context, each factor represents a simple equation that can be set to zero to solve for its root.
When \(2x - 3 = 0\), we solve for \(x\) giving us one root: \(x = \frac{3}{2}\).
Similarly, \(2x + 3 = 0\) provides another root: \(x = -\frac{3}{2}\). Expressing a polynomial as a product of linear factors not only simplifies the solving process but also gives a clear picture of the polynomial's behavior visually.