Problem 21 - Algebra and Trigonometry

Question

13- 30 . Factor the polynomial completely and find all its zeros. State the multiplicity of each zero. $$ P(x)=16 x^{4}-81 $$

Step-by-Step Solution

Answer

Zeros: \( \frac{3}{2}, -\frac{3}{2}, \frac{3i}{2}, -\frac{3i}{2} \), each with multiplicity 1.

Step 1: Identify the form of the expression

The given polynomial is \( P(x) = 16x^4 - 81 \), which is a difference of squares. The difference of squares formula is \( a^2 - b^2 = (a-b)(a+b) \). Here, identify \( a^2 = 16x^4 \) and \( b^2 = 81 \).

Step 2: Apply the difference of squares formula

Rewrite \( 16x^4 \) and \( 81 \) as squares: \((4x^2)^2 - 9^2\). Now, apply the formula: \( 16x^4 - 81 = (4x^2 - 9)(4x^2 + 9) \).

Step 3: Factor further if possible

The factor \( 4x^2 - 9 \) is also a difference of squares: \((2x)^2 - 3^2\), which can be factored further into \((2x - 3)(2x + 3)\). The term \( 4x^2 + 9 \) is not factorable over the real numbers since it results in imaginary roots.

Step 4: Find the zeros of the polynomial

Set each factor equal to zero and solve for \( x \):1. \( 2x - 3 = 0 \) gives \( x = \frac{3}{2} \).2. \( 2x + 3 = 0 \) gives \( x = -\frac{3}{2} \).For \( 4x^2 + 9 = 0 \), solving gives complex zeros \( x = \pm \frac{3i}{2} \).

Step 5: Determine multiplicities of the zeros

Since the factors \((2x - 3)\) and \((2x + 3)\) each appear once, both real zeros \( x = \frac{3}{2} \) and \( x = -\frac{3}{2} \) have a multiplicity of 1. The complex zeros \( x = \pm \frac{3i}{2} \) also each have multiplicity 1.

Key Concepts

Difference of Squares

When factoring polynomials, one useful technique is recognizing patterns within the expression. The "difference of squares" is a key pattern that simplifies factoring significantly. This method is applied when a polynomial is structured as a subtraction between two squared terms, like \(a^2 - b^2\). To factor using the difference of squares, we utilize the formula: \(a^2 - b^2 = (a-b)(a+b)\). This formula allows us to express the polynomial as a product of two binomials. For example, in the polynomial \(16x^4 - 81\), both terms are perfect squares: \( (4x^2)^2\) and \(9^2\). Applying the difference of squares formula, it becomes:

• \( (4x^2 - 9)(4x^2 + 9) \)

This technique is powerful for quickly breaking down expressions into simpler components, revealing their underlying structure.

Polynomial Zeros

Finding the zeros of a polynomial is crucial to understanding its behavior on a graph. The zeros, also known as roots, are the values of \(x\) where the polynomial equals zero. For the polynomial \(P(x) = 16x^4 - 81\), factoring it helps identify these zeros. Once it is broken down to \((2x - 3)(2x + 3)(4x^2 + 9)\), each factor is set to zero to solve for \(x\). This results in:

• Real zeros: \(x = \frac{3}{2}, \; x = -\frac{3}{2}\)
• Complex zeros: \(x = \pm \frac{3i}{2} \)

Finding these zeros helps predict where the polynomial will intersect the x-axis (for real zeros) and provides insight into its behavior and symmetry related to complex roots.

Multiplicity of Roots

Multiplicity of roots in a polynomial refers to how many times a particular zero occurs. This concept helps us understand not only the zeros but also the shape of the graph at those points. If a zero has a multiplicity greater than one, the polynomial will "touch" the x-axis at this zero without crossing it. However, if the multiplicity is one, it typically crosses the x-axis.For \(P(x) = 16x^4 - 81\), each zero \((x = \frac{3}{2}, \; x = -\frac{3}{2}, \; x = \pm \frac{3i}{2})\) occurs with a multiplicity of one. This means that each real zero results in the graph crossing the x-axis at these points, while the complex zeros do not impact the crossings on the real plane since they lie off the real axis. Thus, understanding multiplicity helps visualize the graph's behavior around its zeros.