Problem 73 - A Graphical Approach to Precalculus with Limits

Question

Use the given zero to completely factor \(P(x)\) into linear factors. Zero: \(-2 i ; \quad P(x)=x^{4}+x^{3}+2 x^{2}+4 x-8\)

Step-by-Step Solution

Answer

\((x - 2i)(x + 2i)(x - 1)(x + 2)\) is the complete factorization.

Step 1: Understand the Given Conditions

We are given a polynomial \(P(x) = x^{4} + x^{3} + 2x^{2} + 4x - 8\) and one of its zeros \(-2i\). Since the coefficients of the polynomial are real, the complex root \(-2i\) implies that the conjugate \(2i\) is also a root.

Step 2: Form the Quadratic Factor from Complex Roots

The complex roots \(-2i\) and \(2i\) provide a quadratic factor of the polynomial. This can be written as \((x + 2i)(x - 2i) = x^2 + 4\).

Step 3: Divide the Polynomial by the Quadratic Factor

To find the other factors, divide the original polynomial \(P(x)\) by the quadratic factor \(x^2 + 4\). Use either synthetic division or polynomial division. The quotient will be a quadratic polynomial.

Step 4: Perform Polynomial Division

Perform the division of \(P(x)\) by \(x^2 + 4\). Divide \(x^4 + x^3 + 2x^2 + 4x - 8\) by \(x^2 + 4\) to get \(x^2 + x - 2\) as the quotient without any remainder.

Step 5: Factor the Quotient Polynomial

Now, factor the quotient polynomial \(x^2 + x - 2\). We need to find two numbers that multiply to \(-2\) and add up to \(1\). These numbers are 2 and -1. The factorization is \((x - 1)(x + 2)\).

Step 6: Write the Complete Factorization

Combine the factors to express \(P(x)\) as a product of linear factors. We previously found the quadratic factor \(x^2 + 4\) and the linear factors from the quotient \((x - 1)(x + 2)\). Thus, \(P(x) = (x^2 + 4)(x - 1)(x + 2)\). As \(x^2 + 4\) corresponds to \((x - 2i)(x + 2i)\), the full factorization is \((x - 2i)(x + 2i)(x - 1)(x + 2)\).

Key Concepts

Complex Roots

When dealing with polynomials whose coefficients are real numbers, encountering complex roots often means working with numbers involving the imaginary unit \(i\). In this context, complex roots usually appear as conjugate pairs. For instance, if \(-2i\) is a root of the polynomial, its conjugate \(2i\) must also be a root. This pairing ensures that the resulting coefficients, after expanding their product, remain real numbers.

Here's why that happens:

• The complex conjugate of \(a + bi\) is \(a - bi\). When you multiply \((a + bi)\) and \((a - bi)\), the imaginary parts cancel out creating a real number \(a^2 + b^2\).

Understanding this relationship is key to simplifying the factorization process for polynomials with complex zeroes.

Quadratic Factor

Once you identify a pair of complex roots, like \(-2i\) and \(2i\), you can immediately construct a quadratic factor of your polynomial. For example, given these roots, you combine them into a factor:

• \((x + 2i)(x - 2i)\)
• Utilizing the identity \((a+b)(a-b) = a^2 - b^2\), the quadratic factor becomes \(x^2 + 4\)

The genius of this approach is that these pairs of roots simplify back into terms that are real-number friendly, preserving the solvability of the factorization in the realm of real numbers only.

Polynomial Division

Polynomial division is similar to long division with numbers, but here you work with variables and coefficients. In this exercise, dividing the original polynomial \(P(x) = x^{4} + x^{3} + 2x^{2} + 4x - 8\) by the quadratic factor \(x^2 + 4\) helps us weed out simpler components.

Here's how it works:

• Align the terms in decreasing order of their degrees.
• Take the leading term of the dividend and divide it by the leading term of the divisor to find the first term of the quotient.
• Multiply the entire divisor by this quotient term and subtract from the dividend.
• Repeat the process until you reach a remainder that is lower in degree than the divisor.

This division technique reduces the degree of the polynomial, unearthing another layer of factors, which can be further inspected for more roots.

Linear Factors

When we break down a polynomial completely, the goal is to express it in terms of linear factors. A linear factor is of the form \((x-a)\), where \(a\) is a root of the polynomial. After performing polynomial division on \(P(x)\), we get a quotient of \(x^2 + x - 2\).

Next, we factor the quadratic \(x^2 + x - 2\) into its linear components. We look for two numbers that multiply to \(-2\) and add to \(1\). These numbers are \(2\) and \(-1\). Hence, the quadratic factorizes further as:

• \((x - 1)(x + 2)\)

Combining this with the earlier found complex root factorization, the complete factorization of the original polynomial results in the linear factors:

• \((x - 2i)(x + 2i)(x - 1)(x + 2)\)

This step is crucial as it encapsulates the entire polynomial into its simplest building blocks.