What constant $k$ will make $y^2 + 24y + 16 + k$ the square of a binomial?
I want to write $y^2 + 24y + 16 + k$ as $(y + a)^2$ for some $a$. Expanding $(y + a)^2$, I get $y^2 + 2ay + a^2$. Comparing this with $y^2 + 24y + 16 + k$, I see that $2a = 24$ and $a^2 = 16 + k$. Solving for $a$, I get $a = 12$. Plugging this into $a^2 = 16 + k$, I get $144 = 16 + k$. Solving for $k$, I get $k = 128$. 