Problem 26 - Algebra and Trigonometry

Question

Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. \(P(x)=x^{3}-64\)

Step-by-Step Solution

Answer

The zeros are 4, -2+2i√3, and -2-2i√3, each with multiplicity 1.

Step 1: Recognize the Type of Polynomial

The polynomial \(P(x) = x^3 - 64\) is a difference of cubes because it can be rewritten as \(P(x) = x^3 - 4^3\). A difference of cubes has the form \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\).

Step 2: Rewrite as a Difference of Cubes

Identify \(a = x\) and \(b = 4\). Substitute into the difference of cubes formula: \(x^3 - 64 = (x - 4)(x^2 + 4x + 16)\).

Step 3: Find the Zeros of Each Factor

Set each factor equal to zero and solve. For \(x - 4 = 0\), we get \(x = 4\).\ For the quadratic \(x^2 + 4x + 16 = 0\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 4\), \(c = 16\).

Step 4: Calculate the Discriminant and Solve

Calculate the discriminant \(b^2 - 4ac = 16 - 64 = -48\). Since the discriminant is negative, the quadratic has complex solutions. Use the quadratic formula: \[ x = \frac{-4 \pm \sqrt{-48}}{2} = \frac{-4 \pm i\sqrt{48}}{2} = \frac{-4 \pm 4i\sqrt{3}}{2} = -2 \pm 2i\sqrt{3}. \]

Step 5: State the Zeros and Their Multiplicities

The zeros of the polynomial are \(x = 4\) with multiplicity 1, and \(x = -2 + 2i\sqrt{3}\) and \(x = -2 - 2i\sqrt{3}\) each with multiplicity 1.

Key Concepts

Difference of Cubes

When you see a polynomial like \( P(x) = x^3 - 64 \), you should identify it as a **difference of cubes**. This is a special form of polynomial which can be factored using the formula:

• \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \).

In this case, you can recognize that \( a = x \) and \( b = 4 \) because 64 is equal to \( 4^3 \). This gives us:

• \( x^3 - 64 = (x - 4)(x^2 + 4x + 16) \).

The factor \( x - 4 \) is straightforward. It yields a real zero, \( x = 4 \). The second factor, \( x^2 + 4x + 16 \), is a quadratic that requires further analysis to find its zeros. This step reveals much about the structure of the original polynomial.

Quadratic Formula

Once you have factored out the real part, you need to examine the quadratic portion: \( x^2 + 4x + 16 \). To find the zeros, apply the **quadratic formula**:

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

where \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation. In our case, these values are:

• \( a = 1 \), \( b = 4 \), and \( c = 16 \).

First, calculate the discriminant: \( b^2 - 4ac \). For this quadratic, it is \( 16 - 64 = -48 \), which is negative. A negative discriminant indicates that the quadratic does not have real solutions, but rather complex zeros. Utilize the quadratic formula to find the complex numbers:

• \[ x = \frac{-4 \pm \sqrt{-48}}{2} \].

This introduces a complex number component, indicating that the zeros include an imaginary part.

Complex Zeros

Complex numbers arise naturally when dealing with negative discriminants in the quadratic formula. Here, we solve \( x^2 + 4x + 16 = 0 \) using the quadratic formula and find the discriminant \( -48 \). This is crucial because this negative value leads to an **imaginary component**:

• \( \sqrt{-48} = 4i\sqrt{3} \), where \( i \) is the imaginary unit defined as \( \sqrt{-1} \).

Plug these into the quadratic formula:

• \[ x = \frac{-4 \pm 4i\sqrt{3}}{2} \].

Simplifying this expression, we obtain:

• \( x = -2 \pm 2i\sqrt{3} \).

These are the **complex zeros** of the polynomial \( P(x) = x^3 - 64 \). Complex zeros often come in conjugate pairs, as they do here because one is the "mirror" of the other across the real axis. Each complex zero has its own importance, yielding a deeper understanding of polynomial roots beyond just real numbers.