## Online Course Discussion Forum

### MC I-C Fall 2018 Problem 3.22

Re: MC I-C Fall 2018 Problem 3.22

Compare $x^2 + 8x -32$ and $(x+4)^2 = x^2 + 8x +16$. They are almost the same except for the constant terms. So, we can make $x^2+8x-32$ look similar to $(x+4)^2$. To achieve that we need to have a $+16$ at the end, so we can do $$x^2 + 8x + 16 - 16 -32 = (x+4)^2 - 48.$$ "Completing the square" is the process of adding (and subtracting) $16$ so we can have $(x+4)^2$ in the expression.