What is ( t^2-14t+49 ) equal to?

Correct answer: ( (t-7)^2 )

This trinomial has the pattern of a perfect square. The identity is \\(a^2-2ab+b^2=(a-b)^2\\). In \\(t^2-14t+49\\), the first term is \\(t^2\\), and the last term is \\(49=7^2\\). The middle term is \\(-14t=-2\\times t\\times7\\), so the required form is \\(t^2-2\\times t\\times7+7^2\\). Therefore the expression factors as \\((t-7)^2\\).

Option B is correct. Expanding it confirms the result: \\((t-7)^2=t^2-14t+49\\). Option A would produce a positive middle term, \\(t^2+14t+49\\), so it cannot match. The constants 14 and 49 should be checked through the square identity rather than guessed from the final term alone.