How do I get from $-4(x+\frac{1}{2})^2+4$ to $4-(2x+1)^2$? I'm trying to follow along with printed solutions to previous exam questions, and I'm completely stumped as to how this makes any sense.
I had to complete the square for the function $3-4x-4x^2$, and I got $-4(x+\frac{1}{2})^2+4$. But that's not the final form I need it in because I"m trying to use the solution to integrate a function... so the solution says that I need it in the form $4-(2x+1)^2$, but I cannot figure out how to transform it to that. It could very well be because I'm a bit tired right now, but could somebody point out how it's happening?
 A: *

*You do not need to put it in that form to integrate.  The form you have it in is fine.

*You can use the identities $-a+b=b-a$, and $4a^2=(2a)^2$.  Note that there is no need to redo your completing the square; you can just absorb the $4=2^2$ into the square.

A: $$-4(x+\frac{1}{2})^2 + 4$$
$$-4(x^2 + \frac{1}{2}x + \frac{1}{2}x + \frac{1}{4}) + 4$$
$$-4x^2 - 2x - 2x - 1 + 4$$
$$4 + (-4x^2 - 4x - 1)$$
$$4 - (4x^2 + 4x + 1)$$
$$4 - (2x + 1)^2$$
A: You can use either forms (they are equivalent) to integrate: the solution you are referring to just offers one form of the function for integrating purposes; it's not the only "correct" form of the function.
Applying the suggestions posted by by @Jonas Meyer:
$$
\begin{align} -4\left(x + \dfrac 12\right)^2 + 4 
& \;= \;4 - 4\left( x +\dfrac 12 \right)^2 \\
& =\; 4 - 2^2\left( x +\dfrac 12 \right)^2 \\
& = \;4 - \left(2\left(x + \dfrac 12 \right)\right)^2 \\ \\
& = \;4 - \left(2x+1\right)^2
\end{align}
$$
The process involved in obtaining the given "solution", if you begin from the start, is just another way to "complete the square": 
$$
\begin{align} 3 - 4x - 4x^2 \;
& =\; 4 - 1 - 4x - 4x^2 \\ \\
& =\; 4 - (4x^2 + 4x + 1) \\ \\
& =\; 4 - (2x+1 )^2 \\
\end{align}$$
giving you a function of the form $a^2 - b^2$ where $a = 2$, and $b = (2x+1)$.
Depending on the method used to evaluate the relevant integral, one form may be easier to work with than the other.
A: $-4(x+ \dfrac{1}{2})^2+4$
$=-2 \cdot2(x^2+\dfrac{1}{4}+x)+4$
$=-4x^2-1-4x+4$
$=4-(4x^2+1+4x)$
$=4-(2x+1)^2$
