Problem 16 - Algebra and Trigonometry

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 are 2, -1, \(-\frac{1}{2} \pm \frac{\sqrt{3}}{2}i\); factor is \((x-2)(x+1)(x^2+x+1)(x^2-x+1)\)."

Step 1: Substitute and Simplify P(x)

Given the polynomial \( P(x) = x^6 - 7x^3 - 8 \). Recognize that it can be simplified by setting \( y = x^3 \). Thus, \( P(x) \) becomes \( y^2 - 7y - 8 \). This substitution allows us to treat it as a quadratic equation.

Step 2: Solve the Quadratic Equation

Solve for \( y \) in the equation \( y^2 - 7y - 8 = 0 \) using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -7 \), and \( c = -8 \). Calculate the discriminant \( b^2 - 4ac = 49 + 32 = 81 \). Hence, \( y = \frac{7 \pm \sqrt{81}}{2} = \frac{7 \pm 9}{2} \). This gives \( y = 8 \) and \( y = -1 \).

Step 3: Transform to Original Variable

Recall that \( y = x^3 \). For \( y = 8 \), \( x^3 = 8 \) implies \( x = 2 \). For \( y = -1 \), \( x^3 = -1 \) implies \( x = -1 \). Thus, the roots of \( x^3 = 8 \) are \( x = 2 \), and for the cubic roots of \( -1 \) are \( x = -1, -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i \) based on the cube roots of unity.

Step 4: List All Zeros

The zeros of the polynomial \( P(x) \) are \( 2, -1, -\frac{1}{2} + \frac{\sqrt{3}}{2} i, -\frac{1}{2} - \frac{\sqrt{3}}{2} i \).

Step 5: Factor the Polynomial Completely

Using the zeros identified, the polynomial \( P(x) \) can be expressed in its factored form as \( P(x) = (x-2)(x+1)\left(x + \frac{1}{2} - \frac{\sqrt{3}}{2} i\right)\left(x + \frac{1}{2} + \frac{\sqrt{3}}{2} i\right)\left(x^3 - 8x^3 + 1\right) \). Simplify to group similar roots.

Key Concepts

Complex Numbers

Complex numbers are an essential part of mathematics, especially when dealing with polynomial roots that aren't real numbers. They consist of a real part and an imaginary part. The imaginary unit is denoted as \( i \), where \( i^2 = -1 \). Complex numbers are written in the form \( a + bi \).

• \( a \) is the real part, and
• \( b \) is the imaginary coefficient.

In the given exercise, we encountered complex numbers as solutions when finding the roots of a polynomial. Complex conjugates are a common occurrence when dealing with polynomials. If \( a + bi \) is a root, then \( a - bi \) is often a root as well, ensuring that the polynomial has real coefficients. Understanding these complex solutions is crucial as they provide a complete picture of the polynomial's behavior.

Quadratic Equation

Quadratic equations are fundamental in algebra, having the general form \( ax^2 + bx + c = 0 \). Solving them involves various methods, one of the most famous being the quadratic formula:

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

This formula provides solutions by calculating the roots, determined by the discriminant \( b^2 - 4ac \). In our exercise, an original polynomial was transformed into a quadratic form \( y^2 - 7y - 8 = 0 \) by substituting \( y = x^3 \). Solving this quadratic helps find values for \( y \), which we then relate back to \( x \).
Understanding how to transition between different forms and leverage the quadratic equation is a crucial problem-solving skill.

Factoring Polynomials

Factoring polynomials is a method of expressing the polynomial as a product of its factors, often to solve equations. In the exercise solution, we factor by using known roots, \( a \). If \( a \) is a root, then \( (x - a) \) is a factor of the polynomial.

• Find the roots of the polynomial.
• Use these roots to express the polynomial in factored form.

In this example, roots were derived from solving both the quadratic equation and using properties of cube roots. This leads to expressing the polynomial \( P(x) \) in terms of its linear factors, involving both real and complex terms. Factoring is indispensable for solving polynomial equations entirely and understanding their structure.

Polynomial Equations

Polynomial equations involve variables raised to different powers and combined with coefficients. The highest power of the variable is called the degree of the polynomial and greatly influences the equation's complexity.

• Degree influences the number of possible roots.
• Roots can be real or complex.

In the original exercise, \( P(x) = x^6 - 7x^3 - 8 \) presented a sixth-degree polynomial. By methodical transformations and solving, we found all six roots, both real and complex. Understanding polynomial equations allows you to predict the behavior of the function and determine its intersections with the axis. Advanced topics include polynomial division and using synthetic division, which were not needed here but are useful techniques in solving and simplifying polynomials.