Problem 78 - Algebra and Trigonometry

Question

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 real numbers but can be factored over the complex numbers as \((x - 6i)(x + 6i)\).

Step 1 - Identify the Pattern

The given polynomial \(x^{2} + 36\) can be written as \(x^{2} + 6^{2}\), which is a sum of squares. This fits into the pattern \(a^{2} + b^{2}\) where \(a = x\) and \(b = 6\).

Step 2 - Apply the Formula over the Complex Numbers

If the polynomial were a difference of squares, it could be easily factored over the real numbers. However, sums of squares like \(x^{2} + 6^{2}\) cannot be factored over the real numbers. But, over the complex numbers, sums of squares are factorable using the formula \(a^{2} + b^{2} = (a - bi)(a + bi)\). Here \(i\) is the imaginary unit with the property \(i^{2} = -1\).

Step 3 - Substitute and Simplify

Substitute \(a = x\) and \(b = 6\) into the formula to get: \((x - 6i)(x + 6i)\).