If $x$ is a real number, find $(x+1)^2+2(x+1)(3-x)+(3-x)^2$.
I notice that this expression has the form $a^2+2ab+b^2$, where $a=x+1$ and $b=3-x$. This is a perfect square, so I can rewrite it as $(a+b)^2$. Plugging in the values of $a$ and $b$, I get $(x+1+3-x)^2$. Simplifying the parentheses, I get $(4)^2$. This is just 16. 