Problem 70 - College Algebra

Question

In Exercises \(57-84\), factor completely, or state that the polynomial is prime. $$x^{2}+36$$

Step-by-Step Solution

Answer

The polynomial \(x^{2} + 36\) is prime over the set of real numbers. However, over the set of complex numbers, it can be factorized into \((x + 6i)(x - 6i)\)

Step 1: Identify the form of the expression

The expression \(x^{2} + 36\) can be written as \(x^{2} + 6^{2}\), which is a sum of squares.

Step 2: Attempt Factoring Over Real Numbers

Normally, the factoring of a sum of squares in real numbers isn't possible. Unlike difference of squares, \(a^{2} - b^{2} = (a - b)(a + b)\), the sum of squares, \(a^{2} + b^{2}\), does not factorize over the set of real numbers.

Step 3: Factoring Over Complex Numbers

While one cannot factorize \(x^{2} + 6^{2}\) over the set of real numbers, it is possible over the set of complex numbers. The sum of squares formula over the set of complex numbers is \(a^{2} + b^{2} = (a + ib)(a - ib)\), where \(i = \sqrt{-1}\). Therefore, \(x^{2} + 6^{2} = (x + 6i)(x - 6i)\).

Key Concepts

Sum of Squares

The term "sum of squares" refers to the expression of the form \(a^2 + b^2\). This type of polynomial is quite common in algebraic exercises. While factoring techniques for the **difference of squares** are well-established over real numbers, the **sum of squares** presents unique challenges.

• It can be rewritten in terms of squares, such as \(x^2 + 6^2\) for the expression \(x^2 + 36\).
• Unlike the difference of squares \((a^2 - b^2)\), there is no analogous simple factorization for sum of squares over real numbers.

In real numbers, we say that the sum of squares is "prime," meaning it's not factorizable further. However, when we expand our view to include complex numbers, new possibilities for factoring emerge.

Complex Numbers

Complex numbers are numbers that include the imaginary unit \(i\), where \(i = \sqrt{-1}\). These numbers extend the real number domain and allow us to solve equations that cannot be solved with real numbers alone.

• They are written in the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part.
• In the context of sum of squares, they provide a useful tool because they enable us to factor these expressions completely.

For example, for the expression \(x^2 + 36\), we use complex numbers to rewrite it as \((x + 6i)(x - 6i)\), taking advantage of the imaginary component to achieve the factorization.

Difference of Squares

Unlike the sum of squares, the difference of squares \(a^2 - b^2\) has a straightforward factorization formula: \((a - b)(a + b)\). This property makes it a powerful tool in polynomial factoring.

• This formula works because the cross terms \(-ab\) and \(+ab\) cancel each other out, leaving only the square terms \(a^2\) and \(-b^2\).
• Factorization of differences of squares is possible over real numbers without the need for complex numbers, differentiating it from the sum of squares.

Thus, while the sum and difference of squares both involve squared terms, their algebraic handling in terms of factorization is markedly different.

Real Numbers vs Complex Numbers

Understanding the distinction between real numbers and complex numbers is crucial for mastering polynomial factoring, particularly involving sum and difference of squares.

• Real numbers consist of all the numbers on the continuous number line, including both rational and irrational numbers.
• Complex numbers extend this idea, including a real and an imaginary component, thereby increasing the possible solutions for polynomials.

For example, while the sum of squares \(x^2 + 36\) is not factorable over real numbers, considering complex numbers allows us to express this as \((x + 6i)(x - 6i)\). This understanding helps in tackling a wider range of algebraic problems by utilizing both real and complex number solutions.