Problem 15 - College Algebra

Question

In Exercises \(15-22,\) use the given root to find the solution set of the polynomial equation. $$ x^{3}-2 x^{2}+4 x-8=0 ;-2 i $$

Step-by-Step Solution

Answer

The solution set of the given polynomial is \{-2i, 2i, 2\}

Step 1: Use the complex conjugate root theorem

The complex conjugate root theorem guarantees that if a complex number \(a + bi\) is a root of a polynomial with real coefficients, then its conjugate \(a - bi\) is also a root. In this case, given that \(-2i\) is a root, we can immediately conclude that the conjugate, \(2i\), is also a root.

Step 2: Represent the polynomial in factored form

A polynomial equation of degree \(n\) should have \(n\) roots. Here, given that the polynomial is of degree 3, and we already have 2 roots, there should be one more root. To find this root, we represent the polynomial in its factored form. As we know that \(-2i\) and \(2i\) are roots, the factored form would look like \((x + 2i)(x - 2i)(x - r) = 0\). \('r'\) signifies the third root we need to find.

Step 3: Simplify and find third root

Simplifying \((x + 2i)(x - 2i)\) using difference of two squares results in \(x^{2} + 4\). So now our polynomial equation becomes: \((x^{2} + 4)(x - r) = x^{3} - 2x^{2} + 4x - 8\). Comparing the coefficients of \(x^{3}\), \(x^{2}\), \(x\) and constant term on both sides will help us find the value of 'r'. Upon comparing, we find that \(r = 2\).

Key Concepts

Complex Conjugate Root Theorem

In the realm of polynomial equations, the Complex Conjugate Root Theorem is a valuable tool when dealing with real coefficients. It states:

• If a polynomial has real coefficients and a non-real complex number \(a + bi\) as a root, then the complex conjugate \(a - bi\) must also be included as a root.

This theorem assures that complex roots of polynomials appear in conjugate pairs. For instance, if the polynomial in question has the complex number \(-2i\) as one of its roots, the theorem asserts that its conjugate, \(2i\), is automatically a root.

The presence of conjugate roots helps maintain the integrity of polynomials with real coefficients. This is because the product of a complex number and its conjugate results in a quadratic expression with real coefficients. Therefore, knowing one complex root allows us to easily identify its pair, simplifying the process of factorization and root finding.

Polynomial Equations

Polynomial equations are expressions composed of variables and coefficients, linked by arithmetic operations such as addition, subtraction, multiplication, and non-negative integer exponents on the variables. A general polynomial equation is formatted as:

\[a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0\]

where \(n\) is a non-negative integer representing the degree of the polynomial, and \(a_n, a_{n-1}, \ldots, a_0\) are constants, with \a_n eq 0\.

The degree of the polynomial dictates the maximum number of roots it can have. Therefore, a cubic polynomial, such as \(x^3 - 2x^2 + 4x - 8 = 0\), has up to three roots.

When solving polynomial equations, factoring is a common strategy. Factors of the polynomial represent its roots or solutions. Thus, breaking down the polynomial into its factors can expose all possible solutions, including real and complex roots.

Roots of Polynomials

The roots of a polynomial equation are the values of the variable that satisfy the equation, rendering the polynomial equal to zero. Understanding and finding these roots are pivotal in polynomial analysis.

To determine the roots, consider the polynomial factorization approach. For a cubic polynomial like \(x^3 - 2x^2 + 4x - 8 = 0\), factorization can reveal solutions in terms of simpler polynomial factors:

• If \-2i\ is a root (as given), then according to the complex conjugate root theorem, \2i\ is also a root.
• This insight allows the expression of the polynomial in factored form: \((x + 2i)(x - 2i)(x - r) = 0\), where \(r\) needs to be discovered.

Finding the remaining root involves comparing the factored polynomial with the original polynomial.

By expanding and simplifying using the difference of squares, you determine that \(x^{2} + 4\) is part of the polynomial's factored form. Continuing the process reveals that the third root \r = 2\, completing the solution set. Recognizing this structure and its roots is crucial for understanding the entire behavior of the polynomial.