A Leading Number

An expression has no sides to divide, so when a is not 1 you take a out of the two x terms instead. Take $3{x}^{2}+6x+4$. Taking 3 out of $3{x}^{2}+6x$ gives $3\left({x}^{2}+2x\right)+4$. The number 4 stays outside, untouched.

Now complete the square inside the bracket. The missing number for ${x}^{2}+2x$ is 1, so ${x}^{2}+2x$ is ${\left(x+1\right)}^{2}-1$, and the expression is $3\left({\left(x+1\right)}^{2}-1\right)+4$. Expanding the outer 3 gives $3{\left(x+1\right)}^{2}-3+4$, which is $3{\left(x+1\right)}^{2}+1$.

Watch the expanding step: the 3 multiplies the borrowed -1 as well, so the amount taken back is 3, not 1. That is the most common slip here. The vertex is (-1, 1), and a = 3 says the curve opens up, steeply.