Which expression cannot be factorised into two non-constant polynomials with integer coefficients?

Correct answer: \(m^2+16\)

The direct answer is C: \(m^2+16\) cannot be split into two non-constant polynomials with integer coefficients. The difference-of-squares identity works only for subtraction: \(A^2-B^2=(A-B)(A+B)\). Option A is \(x^2-7^2=(x-7)(x+7)\). Option B is \((5a)^2-(2b)^2=(5a-2b)(5a+2b)\). Option D is \((3p)^2-q^2=(3p-q)(3p+q)\). In C, the expression is a sum of squares, \(m^2+4^2\), not a difference. Over the integers, there is no corresponding identity that produces two non-constant factors. A common confusion is to write \((m-4)(m+4)\), but that product equals \(m^2-16\), with a minus sign, not \(m^2+16\). Always check the sign before applying the difference-of-squares formula.