If $a$ is a constant such that $9x^2 + 24x + a$ is the square of a binomial, then what is $a$?
I want to find a constant $a$ that makes $9x^2 + 24x + a$ a perfect square. That means I should be able to factor it as $(3x + b)^2$, where $b$ is some other constant. Expanding $(3x + b)^2$, I get $9x^2 + 6bx + b^2$. Comparing this to $9x^2 + 24x + a$, I see that $6bx = 24x$, so $b = 4$. Then, $b^2 = 16$, so $a = 16$ as well. To check, I can plug in $a = 16$ and see if I can factor the expression as a perfect square. Indeed, $9x^2 + 24x + 16 = (3x + 4)^2$, so I have found the correct value of $a$. 