Problem 66 - Precalculus Mathematics for Calculus

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. $$P(x)=x^{4}+8 x^{2}+16$$

Step-by-Step Solution

Answer

(a) \((x^2 + 4)^2\), (b) \((x + 2i)^2(x - 2i)^2\).

Step 1: Recognize the Structure of the Polynomial

The polynomial given is a quartic (degree 4) function: \(P(x) = x^4 + 8x^2 + 16\). Notice that this can be seen as a quadratic in terms of \(x^2\): \(P(x) = (x^2)^2 + 8(x^2) + 16\).

Step 2: Substitute to Simplify the Expression

Let \(u = x^2\), then \(P(x)\) becomes a quadratic: \(u^2 + 8u + 16\). We aim to factor this quadratic expression.

Step 3: Factor the Quadratic Polynomial

The expression \(u^2 + 8u + 16\) can be factored as \((u + 4)(u + 4)\) or \((u + 4)^2\), since \(4 + 4 = 8\) and \(4 \times 4 = 16\).

Step 4: Replace \(u\) Back with \(x^2\)

Substitute back \(u = x^2\) in the factored expression to get: \((x^2 + 4)^2\). This gives us the factorization in terms of \(x\) with real coefficients: \((x^2 + 4)^2\). This is the answer to part (a).

Step 5: Factor Each Quadratic Completely Over the Complex Numbers

Each quadratic term \(x^2 + 4\) can be factored using complex numbers as: \((x + 2i)(x - 2i)\).

Step 6: Write the Fully Factored Form with Complex Coefficients

Since \((x^2 + 4)^2 = (x^2 + 4)(x^2 + 4)\), using the factorization from Step 5, we can write: \((x + 2i)(x - 2i)(x + 2i)(x - 2i)\) or \((x + 2i)^2(x - 2i)^2\). This is the answer to part (b).

Key Concepts

Quartic Polynomial

Understanding what a quartic polynomial is can guide you in solving complex problems. A quartic polynomial is a polynomial of degree 4, meaning the highest exponent of the variable is 4. Such polynomials can look intimidating at first, but often there's a simpler underlying structure. In our exercise, the polynomial is:

• \( P(x) = x^4 + 8x^2 + 16 \)

You'll notice this as a special type of quartic polynomial called a "biquadratic," because it can be rewritten as a quadratic in terms of \( x^2 \) instead of \( x \), which simplifies the process of factorization. Recognizing patterns and structures, like seeing that the polynomial is structured as a square \[ (x^2)^2 \], is key to tackling these higher-degree polynomials with ease.
Taking note of polynomial structures not only helps in factorization but also can lead to solving equations and other algebraic manipulations.

Quadratic Substitution

Quadratic substitution is a technique used to make complex polynomial equations more manageable by substituting a part of the polynomial with a simpler variable. In simple terms, you "replace" a part of the equation with another variable, usually denoted as \( u \).

• In the example from the exercise:
  Let \( u = x^2 \), turning the original quartic polynomial into a quadratic: \( u^2 + 8u + 16 \).

This substitution changes the perspective entirely. It simplifies the fourth-degree polynomial into a more familiar second-degree (quadratic) one, making factorization straightforward.
Once the simpler form is factored, you re-substitute the original variable relationship, which in this case means replacing \( u \) back with \( x^2 \). This transforms the expression back to the context of the original problem but in a factorized form.
Quadratic substitution is a powerful technique, especially when dealing with polynomials of even degrees where the exponents can easily be halved.

Complex Factorization

Complex factorization allows you to factor polynomials completely when real coefficients are not sufficient. Often, polynomials like \( x^2 + 4 \) can't be factored using only real numbers because there's no real number \( x \) that satisfies \( x^2 + 4 = 0 \).

• Here, complex numbers are used, specifically imaginary numbers, to express these factors:
  The roots of \( x^2 + 4 = 0 \) are \( x = 2i \) and \( x = -2i \), where \( i \) is the imaginary unit, \( i = \sqrt{-1} \).

With this method, the factorization of \( x^2 + 4 \) becomes \( (x + 2i)(x - 2i) \).
Hence, the full factorization of the original polynomial using complex numbers is: \( (x + 2i)^2(x - 2i)^2 \).
Complex factorization expands the toolkit for solving polynomials, giving solutions and factors which are invisible within the realm of real numbers. This technique is essential in fields such as engineering and physics where complex numbers are regularly employed.