Problem 68 - Algebra and Trigonometry

Question

A polynomial P is given. (a) Factor P into linear and irreducible quadratic factors with real coefficients. (b) Factor P completely into linear factors with complex coefficients. \(P(x)=x^{5}-16 x\)

Step-by-Step Solution

Answer

(a) \(P(x) = x(x - 2)(x + 2)(x^2 + 4)\); (b) \(P(x) = x(x - 2)(x + 2)(x - 2i)(x + 2i)\).

Step 1: Identify Common Factors

First, we observe the polynomial \(P(x) = x^5 - 16x\). Notice that both terms have an \(x\) in common. Factor out \(x\) from the expression: \[P(x) = x(x^4 - 16)\].

Step 2: Recognize Special Form

The expression \(x^4 - 16\) is a difference of squares \((a^2 - b^2)\), where \(a = x^2\) and \(b = 4\). We can apply the difference of squares formula: \[a^2 - b^2 = (a - b)(a + b)\].

Step 3: Apply Difference of Squares

Using the difference of squares, we factor \(x^4 - 16\) as: \[x^4 - 16 = (x^2 - 4)(x^2 + 4)\].Thus, the polynomial is now:\[P(x) = x(x^2 - 4)(x^2 + 4)\].

Step 4: Factor Further

Continue factoring the terms:- \(x^2 - 4\) can be factored further as another difference of squares:\[x^2 - 4 = (x - 2)(x + 2)\].- \(x^2 + 4\) is irreducible over the reals.Thus, the real factorization is:\[P(x) = x(x - 2)(x + 2)(x^2 + 4)\].

Step 5: Factor with Complex Coefficients

The irreducible quadratic \(x^2 + 4\) can be factored over the complex numbers:- Using \(x^2 + 4 = (x - 2i)(x + 2i)\),where \(i\) is the imaginary unit. Thus, the complete linear factorization is:\[P(x) = x(x - 2)(x + 2)(x - 2i)(x + 2i)\].

Key Concepts

Linear Factors

In polynomial factorization, converting a polynomial expression into a product of simpler expressions can help us understand its properties better. When we talk about linear factors, we specifically mean breaking down a polynomial into expressions that are degree one, i.e., of the form \( ax + b \).Linear factors represent the roots of the polynomial, where the polynomial takes the value zero. For the polynomial given, \( P(x) = x^5 - 16x \), we start by factoring out common terms:- Identify common factors: The term "16" and each term in the expression has an \( x \) as a factor, resulting in \( x(x^4 - 16) \).- Further factoring the difference of squares \( x^4 - 16 \) as \((x^2 - 4)(x^2 + 4) \).We find the outright linear factors is possible only for the expression \( x^2 - 4 \) using the difference of squares formula again, simplifying to:

• \( x - 2 \)
• \( x + 2 \)

Thus, within the realm of real numbers, \( P(x) \) primarily simplifies to \( x(x - 2)(x + 2)(x^2 + 4) \). The term \( x^2 + 4 \) cannot be decomposed into linear factors using only real coefficients.

Irreducible Quadratic Factors

In mathematical terms, a quadratic that cannot be factored into linear terms over the set of real numbers is known as an irreducible quadratic factor. This quadratic expression remains as is and cannot be broken down further using real numbers.
Given the polynomial \( P(x) = x(x^2 - 4)(x^2 + 4) \), the expression \( x^2 + 4 \) appears to be a key player.

• While \( x^2 - 4 \) can factor into \( (x - 2)(x + 2) \) due to its nature as a difference of squares,
• the expression \( x^2 + 4 \) remains irreducible over real numbers, as there are no real values of \( x \) for which it equals zero.

In practical purposes, the persistence of this form helps in focusing calculation efforts on what can be manipulated further. It is important to know when a quadratic is irreducible because it directly impacts the ability to express a polynomial solely in linear terms when restricted to real coefficients.

Complex Coefficients

While factoring over real coefficients is quite common, factoring over complex coefficients can provide a more complete view, especially when a quadratic is irreducible over the reals. A polynomial factorized completely means expressed as a product of linear factors, possibly including complex numbers.
For the term \( x^2 + 4 \) which couldn't be split further in the real number system, we turn to complex numbers. This expression can be represented as:

• \( (x - 2i)(x + 2i) \),

where \( i \) is the imaginary unit, satisfying \( i^2 = -1 \).Thus, the polynomial \( P(x) = x^5 - 16x \) can be fully factorized into linear factors incorporating complex numbers:

• \( x, (x - 2), (x + 2), (x - 2i), (x + 2i) \).

This factorization showcases that every polynomial can be reduced to linear factors, although some might require the complex number system.