Problem 74 - Algebra and Trigonometry with Analytic Geometry

Question

Factor the polynomial. $$ 4 x^{2}+9 $$

Step-by-Step Solution

Answer

The polynomial factors as \((2x + 3i)(2x - 3i)\) over the complex numbers.

Step 1: Analyze the Polynomial

The given polynomial is \(4x^2 + 9\). Notice that this is a binomial. The typical strategies to factor binomials involve recognizing them as either a difference of squares, a sum of cubes, or a difference of cubes. However, this expression does not fit any of these forms at first glance, as it is a sum of squares. In the case of real numbers, a sum of squares cannot be factored further using real coefficients.

Step 2: Consider Complex Numbers

Since \(4x^2 + 9\) cannot be factored using real numbers, consider complex numbers. Recall that a sum of squares can sometimes be represented using complex roots. For any sum of squares \(a^2 + b^2\), we can use the fact that \(a^2 + b^2 = (a + bi)(a - bi)\).

Step 3: Applying the Form to the Polynomial

Here \(a^2 = 4x^2\) and \(b^2 = 9\), which implies \(a = 2x\) and \(b = 3\). So, rewrite the expression \(4x^2 + 9 = (2x + 3i)(2x - 3i)\), which are its complex factors.

Key Concepts

Complex Numbers

Complex numbers are an extension of the real numbers, encompassing numbers that have both a real and an imaginary part. The form of complex numbers is generally expressed as \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part, and \( i \) is the imaginary unit \( \sqrt{-1} \).
While complex numbers might sound, well, complex, they actually follow many of the same rules we use with real numbers, such as addition, subtraction, multiplication, and division.

• Addition/Subtraction: Combine like terms. For example, \((a + bi) + (c + di) = (a+c) + (b+d)i\).
• Multiplication: Distribute and simplify using \(i^2 = -1\). For instance, \((a+bi)(c+di) = ac + adi + bci + bdi^2 = (ac-bd) + (ad+bc)i\).
• Division: Apply the conjugate. Dividing by a complex number \(z\) involves multiplying both the numerator and the denominator by the conjugate of \(z\). The conjugate of \(a + bi\) is \(a - bi\).

Understanding complex numbers is crucial when factoring polynomials like \(4x^2 + 9\), which don't factor over the real numbers but do factor when complex numbers are introduced.

Sum of Squares

The sum of squares refers to adding together two perfect squares. Unlike the difference of squares, which can be easily factored, the sum of squares cannot typically be factored into real number coefficients. The expression \(a^2 + b^2\) forms a sum of squares. By introducing complex numbers, however, we find that sum of squares can indeed be factored.
The general formula to factor a sum of squares using complex numbers is \(a^2 + b^2 = (a + bi)(a - bi)\). This expression uses the fact that a sum of squares is equivalent to the product of complex conjugates.
For example, in the polynomial \(4x^2 + 9\), setting \(a=2x\) and \(b=3\) leads us to the factorization \((2x + 3i)(2x - 3i)\), showing how a seemingly unfactorable expression becomes manageable with complex numbers.

Binomials

A binomial is a polynomial expression that consists of exactly two terms. It's an essential concept in algebra and is often encountered in many forms, including being part of larger polynomials.
There are specific methods to factor binomials, some of which are:

• Difference of squares: Expressions such as \(a^2 - b^2\) can be factored as \((a+b)(a-b)\).
• Sum/Difference of cubes: The rules here allow \(a^3 + b^3\) to be rewritten as \((a+b)(a^2-ab+b^2)\), and \(a^3 - b^3\) as \((a-b)(a^2+ab+b^2)\).

In the exercise with the expression \(4x^2 + 9\), we are dealing with a binomial that is a sum of squares. This reflects another key point: binomials don't always follow easy factorization patterns unless you consider extensions like those involving complex numbers. Approaching them with an understanding of complex roots often reveals the hidden factors, broadening the toolkit for polynomial factorization.