Problem 64 - 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. $$ P(x)=x^{5}-16 x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the common factor

First, notice that each term in the polynomial \(P(x) = x^5 - 16x\) includes \(x\). Factor out the greatest common factor, which is \(x\). This gives us: \(P(x) = x(x^4 - 16)\).

Step 2: Recognize the difference of squares pattern

The expression \(x^4 - 16\) can be rewritten as \((x^2)^2 - 4^2\), which is a difference of squares. It can be factored using the identity \(a^2 - b^2 = (a-b)(a+b)\). Apply this to factor \(x^4 - 16\) as \((x^2 - 4)(x^2 + 4)\).

Step 3: Factor further using difference of squares

Notice that \(x^2 - 4\) can further be factored into linear factors since it is also a difference of squares: \(x^2 - 4 = (x - 2)(x + 2)\). Combine this with the previous step to write \(P(x) = x(x - 2)(x + 2)(x^2 + 4)\).

Step 4: Express irreducible quadratic

The factor \(x^2 + 4\) cannot be factored further using real coefficients; hence, it is an irreducible quadratic factor.

Step 5: Factor completely using complex coefficients

The irreducible quadratic \(x^2 + 4\) can be factored into linear factors over complex numbers using the identity \(a^2 + b^2 = (a+bi)(a-bi)\). Thus, \(x^2 + 4 = (x + 2i)(x - 2i)\). Thus, the complete factorization of \(P(x)\) is \(x(x - 2)(x + 2)(x + 2i)(x - 2i)\).

Key Concepts

Difference of Squares

The difference of squares is a pattern that appears frequently in algebraic expressions. It involves expressions of the form \(a^2 - b^2\), which can be factored into \((a-b)(a+b)\). This is useful because it simplifies expressions and helps in solving equations. Recognizing this pattern can make the process of factorization more efficient.

• Example: For an expression like \(x^4 - 16\), notice it can be rewritten as \((x^2)^2 - 4^2\), fitting the difference of squares pattern.
• Factoring: Applying the identity, \(x^4 - 16\) becomes \((x^2 - 4)(x^2 + 4)\).
• Further Factoring: If any of these factors are themselves difference of squares, they can be factored further, such as \(x^2 - 4 = (x-2)(x+2)\).

Being able to spot and apply the difference of squares is a powerful tool in polynomial factorization.

Complex Numbers

Complex numbers extend our concept of real numbers. A complex number has the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2 = -1\).

• Importance in Factorization: Some polynomial expressions that cannot be factored over the real numbers can be factored over the complex numbers. For instance, the expression \(x^2 + 4\) is irreducible over real numbers, but can be decomposed into linear factors involving complex numbers: \((x + 2i)(x - 2i)\).
• Applications: Complex numbers make it possible to fully factorize polynomials that include terms like \(x^2 + k\) where \(k\) is a positive number, since these expressions do not have real roots.

Understanding how to work with complex numbers enhances your ability to completely factor and analyze polynomials.

Quadratic Polynomials

Quadratic polynomials are expressions of the form \(ax^2 + bx + c\). These can often be factored into two linear expressions, provided they have solutions that are real numbers. However, when no real solutions exist, such polynomials become irreducible over the real numbers.

• Irreducible Quadratics: Polynomials like \(x^2 + 4\) do not factor into real linear terms. They remain in their quadratic form if factoring over real numbers.
• Complex Factoring: Though irreducible over the reals, you can factor them over the complex numbers into \((x + 2i)(x - 2i)\), as each complex number solution involves the imaginary unit \(i\).
• Characteristics: The discriminant \(b^2 - 4ac\) in the quadratic formula determines if real, complex, or repeated roots exist. A negative discriminant indicates complex roots.

Mastering quadratic polynomials paves the way for a deeper understanding of more advanced algebraic concepts, especially when considering the field of complex numbers.