What is the correct factorised form of \(y^2+8y+16\)?

Correct answer: \((y+4)^2\)

The direct answer is D: \((y+4)^2\). Begin with the identity \((a+b)^2=a^2+2ab+b^2\). In the given trinomial, the first term is \(y^2\), the last term is \(16=4^2\), and twice the product of y and 4 is \(2\cdot y\cdot4=8y\). Thus \(y^2+8y+16=y^2+2(y)(4)+4^2=(y+4)^2\). Option A gives \(y^2+16y+64\), so its middle and constant terms do not match. Option B gives \(y^2-8y+16\), with the wrong sign in the middle term. Option C gives \(y^2+10y+16\), because the cross terms are 8y and 2y. Option D reproduces all three terms exactly. A useful check is to take the square root of the constant term and then verify the middle term as twice that number times y.