## Algebra 2 Common Core

$y = (x + 2)^{2} -3$
1. $y = x^{2} + 4x + 1$ 2. $y - 1 = x^{2} + 4x$ 3. $y - 1 +$ __ = $x^{2} + 4x +$ __ 4. $(b/2)^{2}= (4/2)^{2} = (2)^{2} = 4$ 5. $y - 1 + 4 = x^{2} + 4x + 4$ 6. $y + 3 = (x + 2)^{2}$ 7. $y = (x + 2)^{2} - 3$ Explanations: 1. This is the beginning equation 2. You must first start by moving C to the other side of the equation. In this case, C is 1, so we moved 1 by subtracting it from both sides. 3. Next, we need to find a number that we can add to both sides that will complete the square. We can put blanks on either side to make it easier for us to remember that we need to add something. 4. In order to find the number needed to complete the square, we can use the formula $(b/2)^{2}$. We plug in b (which is 4) and we solve from there. We put the middle step of $(4/2)^{2}$ because we can use a shortcut and plug that directly into the squared binomial of $(x + 2)^{2}$. We will use this shortcut in step 6. 5. We next use the product of $(b/2)^{2}$ (which was 4) and fill in the blanks from step 3 with it. 6. We then combine like terms on the left side (adding 4 to negative 1). On the right side, we use that shortcut from step 4 and can form the binomial that we need by rewriting $x^{2} + 4x + 4$ as $(x + 2)^{2}$. (you could factor $x^{2} + 4x + 4$ in order to get the binomial, but using the shortcut saves us some time) 7. The last step is to subtract 3 from both sides so that y is by itself. This is your final answer.