Factor completely [tex]$x^2 + 16$[/tex].

A. [tex]\((x + 4)(x + 4)\)[/tex]
B. [tex]\((x + 4)(x - 4)\)[/tex]
C. Prime
D. [tex]\((x - 4)(x - 4)\)[/tex]



Answer :

To determine if the expression [tex]\(x^2 + 16\)[/tex] can be factored completely, let's analyze it step-by-step:

1. Observation of the Form:
The given expression [tex]\(x^2 + 16\)[/tex] is a polynomial in the form of [tex]\(x^2 + c\)[/tex], where [tex]\(c\)[/tex] is a positive constant (16 in this case).

2. Factorization Check:
- We know that certain types of expressions have specific factorization formulas. For instance, the difference of squares is [tex]\(a^2 - b^2 = (a + b)(a - b)\)[/tex], and perfect square trinomials like [tex]\(a^2 + 2ab + b^2 = (a + b)^2\)[/tex].
- However, [tex]\(x^2 + 16\)[/tex] is not a difference of squares; instead, it is a sum of squares.

3. Sum of Squares:
- Unlike the difference of squares, the sum of squares does not have a factorization over the real numbers in terms of real coefficients. In other words, [tex]\(x^2 + c\)[/tex] cannot be factored further into polynomials with real coefficients unless we allow for complex numbers.

4. Conclusion:
Since [tex]\(x^2 + 16\)[/tex] does not fit any recognizable factorization patterns and cannot be factored into real polynomials, it is considered a prime polynomial.

Therefore, the expression [tex]\(x^2 + 16\)[/tex] is not factorable over the real numbers and is thus Prime.

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