Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$? I'm confused about a question I posted this morning.
I am trying to understand if $-6(x-\frac{1}{2})^2$ can be rewritten as $-\frac{3}{2}(2x-1)^2$?
I tried multiplying out the expression $-6(x-\frac{1}{2})^2$ to a polynomial form $36x^2-36x+9$ but that didn't take me closer to understanding my goal.
I noticed that I can remove the fraction inside $-6(x-\frac{1}{2})^2$ by doubling the contents:
$-6(x-\frac{1}{2})^2$ <> $-6(2x-1)^2$ # used <> for does not equal
I don't think I can simply half the factor -6 to get $-3(2x-1)^2$
As you can no doubt see, I am confused.
Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$ ?
If it does could someone show me how to transform from $-6(x-\frac{1}{2})^2$ to $-\frac{3}{2}(2x-1)^2$ in granular baby steps?
 A: We have
$$-6\left(x-\frac{1}{2}\right)^2=-\frac 6{\color{blue}4}\cdot \color{blue}4\cdot\left(x-\frac{1}{2}\right)^2=-\frac 3{\color{blue}2}\cdot \color{blue}{2^2}\cdot\left(x-\frac{1}{2}\right)^2=\\=-\frac 32\left(\color{blue}2\cdot x-\color{blue}2\cdot\frac{1}{2}\right)^2=-\frac 32(2x-1)^2$$
A: Look at my answer, it's quite simple to prove
$$-6\left(x-\frac{1}{2}\right)^2=-6\left(\frac{2x-1}{2}\right)^2=-6\frac{\left(2x-1\right)^2}{2^2}=-6\frac{\left(2x-1\right)^2}{4}=-\frac 32(2x-1)^2$$
A: $$-6\left(x-\frac{1}{2} \right)^2 = -\frac{3}{2}\left(2x-1 \right)^2$$
$$-6\left( x^2 -x +\frac{1}{4}\right)  = -\frac{3}{2} \left(4x^2-4x+1  \right)$$
$$-6x^2+6x - \frac{6}{4} = -6x^2+6x -\frac{3}{2} $$
A: Your confusion stems from the fact that once you factor out the $\frac{1}{2}$ (aka double the inside), you still need to square it in order to take it outside the parenthesis. So you are correct in what you are trying to do, but then $(\frac{1}{2})^2 = \frac{1}{4}$ which, when multiplied by $-6$, does give you $-\frac{3}{2}$.
