Problem 16 - Precalculus Mathematics for Calculus

Question

A polynomial \(P\) is given. (a) Find all zeros of \(P\), real and complex. (b) Factor \(P\) completely. $$P(x)=x^{6}-7 x^{3}-8$$

Step-by-Step Solution

Answer

Zeros: \(x = 2, -1, -1 \pm i\sqrt{3}, \frac{1}{2} \pm i\frac{\sqrt{3}}{2}\). Factor: \((x - 2)(x + 1)\text{(complex factors)}\).

Step 1: Substitute Variable

Let's begin by substituting the variable to simplify the problem. Use the substitution \(y = x^3\). Then our polynomial becomes \(P(y)=y^2 - 7y - 8\).

Step 2: Solve the Quadratic Equation

Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve \( y^2 - 7y - 8 = 0\). Here, \(a = 1\), \(b = -7\), and \(c = -8\). Calculate:\[y = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm \sqrt{81}}{2} = \frac{7 \pm 9}{2}\]The solutions are \( y = 8 \) and \( y = -1 \).

Step 3: Back-substitution to Original Variable

Reverse the substitution. Since \(y = x^3\), find \(x\) by solving \(x^3 = 8\) and \(x^3 = -1\). The solutions are:- For \(x^3 = 8\): \(x = 2\) (real) and two non-real roots (using complex cube roots of unity) \(-1 + i\sqrt{3}\) and \(-1 - i\sqrt{3}\).- For \(x^3 = -1\): \(x = -1\) (real) and two non-real roots (using complex cube roots of unity) \(\frac{1}{2} + i\frac{\sqrt{3}}{2}\) and \(\frac{1}{2} - i\frac{\sqrt{3}}{2}\).

Step 4: List All Zeros

The polynomial \(P(x)\) has the following roots: real roots \(x = 2\), \(x = -1\) and complex roots \(x = -1 + i\sqrt{3}\), \(x = -1 - i\sqrt{3}\), \(x = \frac{1}{2} + i\frac{\sqrt{3}}{2}\), \(x = \frac{1}{2} - i\frac{\sqrt{3}}{2}\).

Step 5: Factor the Polynomial

Using the roots obtained, factor \(P(x)\). As the roots satisfy the polynomial, the factors are:\[P(x) = (x - 2)(x + 1)(x + 1 - i\sqrt{3})(x + 1 + i\sqrt{3})(x - \frac{1}{2} - i\frac{\sqrt{3}}{2})(x - \frac{1}{2} + i\frac{\sqrt{3}}{2})\]

Key Concepts

Complex Numbers

When dealing with polynomials, you often encounter numbers that are not real. These are called complex numbers. A complex number has a real part and an imaginary part, usually written in the form of \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit. The imaginary unit \( i \) has the property that \( i^2 = -1 \).

Working with complex numbers allows us to find solutions to equations that have no real solutions. For example, when solving a polynomial that appears to have

• a negative discriminant in the quadratic formula,
• or when dealing with the roots of unity.

Understanding complex numbers is crucial for fully solving some polynomial equations since they provide a more complete set of solutions.

Factorization

Factorization involves writing a polynomial as a product of its simpler polynomials, often using its roots. To factor a polynomial completely means to express it as a product of linear terms, possibly including complex numbers.

For example, if you have found the roots of a polynomial, say \( y \) and \( z \), the polynomial can be factored as \((x - y)(x - z)\).

• This bears significance especially when the roots are complex, requiring you to deal with terms involving \( i \).
• In our case, the step-by-step solution factors the polynomial using both real and complex roots into linear terms.

The idea is to express the polynomial in terms of its roots, making it apparent that each root sets one of these factors to zero. This process simplifies further manipulation or understanding of the polynomial.

Quadratic Equation

The quadratic equation is a critical element when breaking down polynomials into simpler parts. It takes the form \( ax^2 + bx + c = 0 \), and its solutions are found using the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \(a\), \(b\), and \(c\) are coefficients from the polynomial.

In the substitution step, the original polynomial is reduced to a quadratic form by substituting \( y = x^3 \). This transformation enables the use of the quadratic formula to find the roots of the transformed equation, further facilitating the process of identifying the original polynomial's roots.

• The quadratic formula is versatile, handling both real and complex roots.
• It's particularly useful when directly factoring a higher degree polynomial cannot be done easily.

Cube Roots of Unity

Cube roots of unity are solutions to the equation \( x^3 = 1 \). They are particularly interesting because they repeat after a full cycle.

• The three cube roots of unity are: \( 1 \), \( \frac{-1 + i\sqrt{3}}{2} \), and \( \frac{-1 - i\sqrt{3}}{2} \).
• They can be derived from a known concept called Euler's formula, which connects complex exponentials and trigonometric functions.

In the context of the polynomial problem, we utilize variations of cube roots (

• such as \( x^3 = 8 \) or \( x^3 = -1 \)

) to determine complex solutions.

Understanding cube roots of unity helps solve cubic equations by acknowledging these repetitive patterns, especially when dealing with complex numbers. This knowledge enables the factorization of polynomials into simpler linear forms, considering both real and complex roots.