Sure! Let's identify the factors of the given expression [tex]\( x^2 + 36y^2 \)[/tex].
To find the factors, we need to determine if the expression can be written as a product of two binomials. The expression [tex]\( x^2 + 36y^2 \)[/tex] resembles a sum of squares, and generally, sums of squares do not factor nicely over the real numbers. Let us explore it step by step:
1. Let's identify the structure of [tex]\( x^2 + 36y^2 \)[/tex].
- [tex]\( x^2 \)[/tex] is the square of [tex]\( x \)[/tex].
- [tex]\( 36y^2 \)[/tex] is the square of [tex]\( 6y \)[/tex] because [tex]\( (6y)^2 = 36y^2 \)[/tex].
2. In order to factor [tex]\( x^2 + 36y^2 \)[/tex], we might consider well-known factoring formulas. Recall, sums of squares normally do not factor over the real numbers. There are special cases like the difference of squares: [tex]\( a^2 - b^2 = (a + b)(a - b) \)[/tex]. But here, we have a sum [tex]\( x^2 + 36y^2 \)[/tex], not a difference.
3. Due to this sum of squares' nature, there is no factorization into real-number binomials. The choices provided in the options do not apply in this case since none satisfy the form of a sum of squares:
- Prime: This means the expression can't be factored using real numbers.
- [tex]\((x + 6y)(x - 6y)\)[/tex]: This applies to differences of squares like [tex]\( x^2 - (6y)^2 = (x + 6y)(x - 6y) \)[/tex], which isn't our case.
- [tex]\((x + 6y)(x + 6y)\)[/tex]: This implies a perfect square of the form [tex]\((a + b)^2\)[/tex], leading to [tex]\( a^2 + 2ab + b^2 \)[/tex], which is also incorrect for our expression.
- [tex]\((x - 6y)(x - 6y)\)[/tex]: Same logic as above for squares.
Given all these analyses, we deduce that [tex]\( x^2 + 36y^2 \)[/tex] does not factorize using real numbers. The expression is already in its simplest form, and thus, we consider it prime for practical purposes.
Therefore, the correct answer is:
- Prime
