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.
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.
