Problem 15 - Precalculus Mathematics for Calculus

Question

A polynomial \(P\) is given. (a) Find all zeros of \(P\), real and complex. (b) Factor \(P\) completely. $$P(x)=x^{6}-1$$

Step-by-Step Solution

Answer

Zeros are 1, -1, and \(\text{cis}(\pi/3), \text{cis}(2\pi/3), \text{cis}(4\pi/3), \text{cis}(5\pi/3)\). The factorization is \((x-1)(x+1)(x^2+x+1)(x^2-x+1)\).

Step 1: Understand the Problem

We are given the polynomial \(P(x) = x^6 - 1\). We need to find its zeros and factor it completely. The polynomial is of degree 6, which suggests it may have complex roots.

Step 2: Find Real Zeros

Observe that the polynomial can be rearranged as \(x^6 = 1\). The real solution is \(x = 1\) and \(x = -1\) because \(1^6 = 1\) and \((-1)^6 = 1\).

Step 3: Find All Complex Roots using De Moivre's Theorem

Recognize that \(x^6 = 1\) is equivalent to finding the 6th roots of unity, given by \(x = e^{2\pi ik/6}\) for \(k = 0, 1, 2, 3, 4, 5\). Calculate these as follows: - \(x = e^{0} = 1\)- \(x = e^{i\pi/3} = \text{cis}(\pi/3)\)- \(x = e^{i2\pi/3} = \text{cis}(2\pi/3)\)- \(x = e^{i\pi} = -1\)- \(x = e^{i4\pi/3} = \text{cis}(4\pi/3)\)- \(x = e^{i5\pi/3} = \text{cis}(5\pi/3)\) where \( \text{cis}(\theta) = \cos(\theta) + i\sin(\theta) \).

Step 4: Express Roots in Simplified Form

Convert each complex exponential into algebraic form: - \(x = 1\)- \(x = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)- \(x = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)- \(x = -1\)- \(x = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\)- \(x = \frac{1}{2} - i\frac{\sqrt{3}}{2}\)

Step 5: Factor the Polynomial Completely

Since \(x^6 - 1 = (x^3 - 1)(x^3 + 1)\), we can factor further: - \((x^3 - 1) = (x - 1)(x^2 + x + 1)\)- \((x^3 + 1) = (x + 1)(x^2 - x + 1)\)This gives the complete factorization: \[P(x) = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1)\]

Step 6: Write the Zeros from Factorization

From the factorization, zeros are 1, -1, and complex roots from \(x^2 + x + 1\) and \(x^2 - x + 1\). They match our previously calculated roots.

Key Concepts

Complex Roots

Polynomials can have both real and complex roots. Complex roots occur when a polynomial equation has no real number solutions. Instead, the solutions can be expressed in the form of complex numbers, which are numbers that include the imaginary unit, denoted as \(i\), where \(i^2 = -1\). For the polynomial \(P(x) = x^6 - 1\), after finding the real roots, we explore complex roots for a complete solution.

To find the complex roots of \(x^6 = 1\), we use the exponential form of complex numbers. We rewrite the problem as \(x = e^{2\pi ik/6}\), where \(k\) is an integer ranging from 0 to 5. This is derived from Euler's formula \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\), indicating that complex numbers can also be represented in their exponential form.

• For \(k=0\), \(x = e^0 = 1\).
• For \(k=1\), \(x = e^{i\pi/3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}\).
• For \(k=2\), \(x = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\).
• For \(k=3\), \(x = e^{i\pi} = -1\).
• For \(k=4\), \(x = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\).
• For \(k=5\), \(x = e^{i5\pi/3} = \frac{1}{2} - i\frac{\sqrt{3}}{2}\).

These complex roots appear in conjugate pairs, which is typical for polynomials with real coefficients.

Zeros of Polynomial

The zeros of a polynomial are the values of \(x\) for which the polynomial evaluates to zero. In other words, they are the solutions to the equation formed by setting the polynomial equal to zero. For \(P(x) = x^6 - 1\), we aim to find both real and complex zeros.

From the factorization step in the solution, we derived the equation as \((x^3 - 1)(x^3 + 1)\). Breaking these further, we get:

• \(x^3 - 1 = (x - 1)(x^2 + x + 1)\)
• \(x^3 + 1 = (x + 1)(x^2 - x + 1)\)

The zeros found by setting each factor to zero are:

• \(x = 1\) from \(x - 1\)
• \(x = -1\) from \(x + 1\)

The zeros from the quadratic factors \(x^2 + x + 1 = 0\) and \(x^2 - x + 1 = 0\) are complex numbers. Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), these were calculated as detailed in the complex roots section.
Thus, the complete list of zeros, confirmed by both the original polynomial and its factored form, is \(1, -1, \frac{1}{2} + i\frac{\sqrt{3}}{2}, -\frac{1}{2} + i\frac{\sqrt{3}}{2}, -\frac{1}{2} - i\frac{\sqrt{3}}{2}, \frac{1}{2} - i\frac{\sqrt{3}}{2}\). This includes both real and complex solutions.

Roots of Unity

Roots of unity are special types of roots that occur when solving the equation \(x^n = 1\). In our case, we looked at the 6th roots of unity for the polynomial \(x^6 - 1\). These are the complex number solutions to the equation and they are evenly distributed points on the complex plane, forming a regular hexagon centered at the origin with a radius of one.
To find the roots of unity, we use the formula \(x = e^{2\pi ik/6}\), which gives us six solutions when \(k = 0, 1, 2, 3, 4, 5\). These correspond to different points on the unit circle in the complex plane:

• \(x = 1\) represents one of these roots.
• The complex roots as calculated earlier, \(\pm \frac{1}{2} + i\frac{\sqrt{3}}{2}\) and \(\pm \frac{1}{2} - i\frac{\sqrt{3}}{2}\), are also roots of unity.
• \(x = -1\), another important root, lies directly opposite \(x = 1\) on the circle.

These 6th roots of unity highlight a fundamental aspect of complex numbers in polynomial factorization. They are equally spaced around the circle and represent the solutions to many problems in mathematics and science, particularly in signal processing and other fields where waveforms and cycles are studied.