How do you factor this using complete the square? $6+12y-36y^2$ I'm so embarrassed that I'm stuck on this simple algebra problem that is embedded in an integral, but I honestly don't understand how this is factored into $a^2-u^2$
Here are my exact steps:
$6+12y-36y^2$ can be rearranged this way: $6+(12y-36y^2)$ and I know I can factor out a -1 and have it in this form: $6-(-12y+36y^2)$
This is the part where I get really lost.  According to everything I read, I take the $b$ term, which is $-12y$ and divide it by $2$ and then square that term.  I get: $6-(36-6y+36y^2)$
The form it should look like, however, is $7-(6y-1)^2$
Can you please help me to understand what I'm doing wrong?
 A: we know that $$ (ay - b)^2 = a^2y^2 - 2ayb + b^2 \space \space \space \space (1) $$
we have $$ 6 + 12y - 36y^2 = -(36y^2 - 12y - 6) $$
we need to factor $$ 36y^2 - 12y - 6 $$
from (1) let $ a^2 = 36 \Rightarrow a = \pm 6, 2ab = 12 \Rightarrow b = \pm 1 $
thus $$ 36y^2 - 12y + 1 - 7  = (6y - 1)^2 \ or \ (-6y +1)^2 - 7 \Rightarrow  6 + 12y - 36y^2 = 7 - (6y-1)^2 $$
A: If you can't recover the algebraic steps of the square completion, you can also do it geometrically. I always do it algebraically, but some students feel more comfortable with the $(h,k)$ approach below. Just an alternative.
The square completion of a quadratic function is nothing but
$$p(x)=ax^2+bx+c=a(x-h)^2+k$$
where $(h,k)$ is the vertex of the parabola given by
$$
h=-\frac{b}{2a}\qquad k=p(h)
$$
Therefore the only thing to remember is $h=-b/2a$ which is the abscissa of the symmetry axis of the parabola, and also the lhs of the quadratic formula: so you already know it.
Try it in your case:
$$a=-36\;b=12\; c=6\Rightarrow h=-\frac{12}{-72}=\frac{1}{6} \Rightarrow k=p(1/6)=-36\cdot \frac{1}{6^2}+12\cdot \frac{1}{6}+6=7$$
whence
$$
p(x)=-36x^2+12x+6=-36\left(x-\frac{1}{6} \right)^2+7 
$$
which is of course equal to $-(6x-1)^2+7$.
A: Here is the first step:
$$6+12y-36y^2 = -\left((6y)^2-2\cdot (6y)-6\right)$$
A complete square would be
$$\left(6y-1\right)^2 = (6y)^2-2\cdot (6y) + 1.$$
Ignoring the overall minus for a moment and adding and subtracting 1 to the r.h.s. of the first equation above, we get (using $a^2-b^2=(a-b)(a+b)$ formula:
$$(6y)^2-2\cdot (6y) - 6 = \left(6y-1\right)^2 - 7 = \left(6y-1 - \sqrt{7}\right) \left(6y-1+\sqrt{7}\right)$$
And finally, restoring the overall minus gets us the answer:
$$6+12y-36y^2 = - \left(6y-1 - \sqrt{7}\right) \left(6y-1+\sqrt{7}\right).$$
