<s>[INST] 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?
 [/INST] $$-6\left(x-\frac{1}{2} \right)^2 = -\frac{3}{2}\left(2x-1 \right)^2$$
$$-6\left( x^2 -x +\frac{1}{4}\right)  = -\frac{3}{2} \left(4x^2-4x+1  \right)$$
$$-6x^2+6x - \frac{6}{4} = -6x^2+6x -\frac{3}{2} $$
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