Problem 25 - Algebra and Trigonometry

Question

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

Step 1: Recognize a Difference of Squares

Notice that the polynomial \( P(x) = 16x^4 - 81 \) can be recognized as a difference of squares. The general formula for a difference of squares is \( a^2 - b^2 = (a-b)(a+b) \). Here, we can rewrite \( 16x^4 \) as \((4x^2)^2\) and \( 81 \) as \(9^2\).

Step 2: Apply the Difference of Squares Formula

Using the difference of squares formula, factor the expression: \[ P(x) = (4x^2 - 9)(4x^2 + 9) \]

Step 3: Factor Completely the Difference of Squares in the Complex Domain

The expression \(4x^2 - 9\) is again a difference of squares and can be factored further:\[ 4x^2 - 9 = (2x - 3)(2x + 3) \]The term \(4x^2 + 9\) is a sum of squares and cannot be factored further using real numbers.

Step 4: Combine the Factors

Combine all the factors obtained:\[ P(x) = (2x - 3)(2x + 3)(4x^2 + 9) \]

Step 5: Find the Zeros of the Polynomial

Setting each factor equal to zero, we find the zeros:- For \(2x - 3 = 0\), solve to get \(x = \frac{3}{2}\).- For \(2x + 3 = 0\), solve to get \(x = -\frac{3}{2}\).The factor \(4x^2 + 9 = 0\) leads to:\[ 4x^2 = -9 \Rightarrow x^2 = -\frac{9}{4} \]This gives complex solutions \(x = \pm \frac{3i}{2}\).

Step 6: Determine the Multiplicity of Each Zero

The zeros \(x = \frac{3}{2}\) and \(x = -\frac{3}{2}\) each occur once, giving them a multiplicity of 1 each. The complex zeros \(x = \frac{3i}{2}\) and \(x = -\frac{3i}{2}\) also have a multiplicity of 1 each.

Key Concepts

Difference of Squares

The concept of a difference of squares is crucial when dealing with polynomial factorization. It allows us to simplify expressions that look complex at first glance. The difference of squares formula is given by \( a^2 - b^2 = (a-b)(a+b) \). This pattern occurs when you have two perfect squares separated by a subtraction sign. Recognizing these patterns can make the process of simplifying polynomials much easier.

In the exercise, the polynomial \( P(x) = 16x^4 - 81 \) becomes a candidate for the difference of squares, because:

• \( 16x^4 \) is the square of \( (4x^2) \)
• \( 81 \) is the square of \( 9 \)

Therefore, it can be expressed using the formula: \( (4x^2)^2 - 9^2 \).
By applying the difference of squares, we break it down to \( (4x^2 - 9)(4x^2 + 9) \), simplifying the expression significantly.

Complex Solutions

Sometimes, polynomials have solutions that include imaginary numbers. These are called complex solutions. As per our differences of squares step, we found \(4x^2 + 9\), which cannot be simplified into real factors. To find its zeros, set the expression equal to zero: \(4x^2 + 9 = 0\).

Following through:

• \(4x^2 = -9\)
• Divide by 4: \(x^2 = -\frac{9}{4}\)
• Take the square root (considering the imaginary unit \(i\)) gives \(x = \pm \frac{3i}{2}\)

Remember, the imaginary unit \(i\) houses the core property that \(i^2 = -1\). Complex solutions are important as they provide complete solution sets for certain polynomials that don’t cross the x-axis.

Zeros of Polynomials

Zeros of a polynomial, often called roots, are the values of \(x\) that make the polynomial equal to zero. To find these, set each factor of the polynomial equation \(P(x) = 0\) and solve for \(x\).

From our exercise, the factors of \(P(x)\) were:

• \(2x - 3 = 0\), yielding \(x = \frac{3}{2}\)
• \(2x + 3 = 0\), yielding \(x = -\frac{3}{2}\)
• The complex component \(4x^2 + 9 = 0\), gives \(x = \pm \frac{3i}{2}\)

These zeros give us a full view of the polynomial's behavior. Each zero corresponds to a point where the polynomial "touches" or "crosses" the x-axis in the graph absis view.

Multiplicity of Zeros

Multiplicity refers to how many times a particular zero appears in the polynomial. This is determined by the number of times a factor is repeated in the factorized form of the polynomial.

In our exercise:

• The zero \(x = \frac{3}{2}\) comes from \(2x - 3\)
• The zero \(x = -\frac{3}{2}\) comes from \(2x + 3\)
• The zeros \(x = \pm \frac{3i}{2}\) are derived from \(4x^2 + 9\)

Each of these zeros occurs only once in the factorization of \(P(x)\), each having a multiplicity of 1. This indicates that the polynomial graph "crosses" at each real zero. Multiplicity helps us understand the graph's nature at these points.