jmc

We will complete the square for $y^2+10y+33.$

The binomial to be squared will be of the form $y+a$ because the coefficient of $y^2$ is 1. By squaring the binomial, we get $y^2+2ay+a^2$. We want $2ay$ to be equal to $10y$, therefore $a=5$. $(y+5{)}^2=y^2+10y+25$.

$y^2+10y+33=(y^2+10y+25)+8=(y+5{)}^2+8$. Therefore, the binomial is $y+5$ and the integer is $\boxed{8}$.