Problem 61 - Algebra and Trigonometry

Question

\(59-64=\) 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. $$ P(x)=x^{4}+8 x^{2}-9 $$

Step-by-Step Solution

Answer

(a) \((x^2 + 9)(x - 1)(x + 1)\); (b) \((x - 1)(x + 1)(x + 3i)(x - 3i)\).

Step 1: Identify the Polynomials

We start with the given polynomial \( P(x) = x^4 + 8x^2 - 9 \). Our task is to factor it into linear and irreducible quadratic factors with real coefficients first, and then fully into linear factors with complex coefficients.

Step 2: Substitute to Simplify

Let \( y = x^2 \). Then \( P(x) = x^4 + 8x^2 - 9 \) becomes a quadratic equation in terms of \( y \): \( y^2 + 8y - 9 \).

Step 3: Factor the Quadratic Equation

Look for two numbers that multiply to \(-9\) and add to \(8\). These numbers are \(9\) and \(-1\). Thus, factor the equation as \((y + 9)(y - 1) = 0\). Substitute back \( y = x^2 \), we have \((x^2 + 9)(x^2 - 1)\).

Step 4: Factor Further with Real Coefficients

The factor \( x^2 - 1 \) can be factored further into \((x - 1)(x + 1)\), since it’s a difference of squares. Thus, \( P(x) = (x^2 + 9)(x - 1)(x + 1) \). Here, \( x^2 + 9 \) is already irreducible over the real numbers.

Step 5: Factor Completely with Complex Coefficients

To factor \( x^2 + 9 \) into linear factors with complex coefficients, recognize it as a sum of squares: \( x^2 + 9 = x^2 + 3^2 = (x + 3i)(x - 3i) \). So, the complete factorization over the complex numbers is \( P(x) = (x - 1)(x + 1)(x + 3i)(x - 3i) \).

Key Concepts

Irreducible Quadratic Factors

When a polynomial is broken down to its basic components with real coefficients, some factors may remain unfactorable. These are called irreducible quadratic factors. In the context of real numbers, a quadratic expression is irreducible if it cannot be factored into two linear factors each having real coefficients. For example, consider the expression \(x^2 + 9\).

• The factorization process is complete when no real numbers fulfill the criteria for breaking this down further as \(x^2 + 9=(x+a)(x+b)\) where \(a\) and \(b\) are real.
• In terms of the given polynomial \(P(x)\) with real coefficients, \(x^2 + 9\) remains an irreducible quadratic factor.

Recognizing irreducible quadratics is essential when working within the realm of real coefficients.

Complex Coefficients

Complex coefficients involve terms with imaginary numbers. This becomes crucial when a factorization that cannot be achieved with real number coefficients is required. For every term that appears irreducible over the reals, such as \(x^2 + 9\), complex numbers step in to decompose them.
Just to recap:

• With real coefficients, \(x^2 + 9\) cannot be broken further. But with complex coefficients, it becomes \((x + 3i)(x - 3i)\), thanks to the imaginary unit \(i\), where \(i = \sqrt{-1}\).
• This reveals the roots, or solutions, as \(x = 3i\) and \(x = -3i\), completing the factorization with complex terms.

By introducing complex coefficients, the polynomial is expressed entirely in terms of linear factors, even those impossible to express through real numbers alone.

Difference of Squares

The difference of squares is a formula in algebra that allows quick factorization. It's stated as \(a^2 - b^2 = (a - b)(a + b)\). When referenced in the polynomials, it is often seen in expressions that can be rewritten as the subtraction of two perfect squares.
Consider the polynomial component \(x^2 - 1\), which fits this structure:

• Here, \(a^2\) is determined as \(x^2\) and \(b^2\) as \(1^2\), allowing it to be factored into \((x - 1)(x + 1)\).
• This factorization step is part of breaking \(P(x)\) into its most reduced form using real coefficients.

This concept is very efficient for simplifying polynomial expressions and helps visually identify terms that can be easily factored, making polynomial manipulation more straightforward.