Factor the biquadratic expression x^4 + 7x^2 + 16 over the reals.

Introduction / Context:
Biquadratic expressions in x^4 can often be factored by treating x^2 as a single variable t. The goal is to express x^4 + 7x^2 + 16 as a product of two real quadratics with zero x^3 and x terms so that expansion matches the original middle term 7x^2 and constant 16.

Given Data / Assumptions:

• Expression: x^4 + 7x^2 + 16
• We seek a factorization over the reals.

Concept / Approach:
Assume a symmetric factorization: (x^2 + ax + 4)(x^2 + bx + 4). To eliminate x^3 and x terms, choose b = −a. Then the x^2 coefficient becomes (ab + 8) = (−a^2 + 8). Match this to 7 to solve for a.

Step-by-Step Solution:
Let factors be (x^2 + ax + 4)(x^2 − ax + 4)Expand: x^4 + (ab + 8)x^2 + 16 with ab = −a^2Coefficient of x^2: −a^2 + 8 must equal 7 ⇒ a^2 = 1 ⇒ a = ±1Thus a = 1 gives (x^2 + x + 4)(x^2 − x + 4)

Verification / Alternative check:
Expand (x^2 + x + 4)(x^2 − x + 4): you get x^4 + 7x^2 + 16 exactly, with x^3 and x coefficients canceling due to symmetry.

Why Other Options Are Wrong:
Other factorizations either create nonzero x^3/x terms or yield incorrect x^2 coefficients/constant terms; “Irreducible over reals” is incorrect since a correct real factorization exists.

Common Pitfalls:
Factoring as a quadratic in x^2 only (which has negative discriminant) and concluding irreducible; but symmetric real quadratic factors still exist.

Final Answer:
(x^2 + x + 4)(x^2 - x + 4)