<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] We have
$$-6\left(x-\frac{1}{2}\right)^2=-\frac 6{\color{blue}4}\cdot \color{blue}4\cdot\left(x-\frac{1}{2}\right)^2=-\frac 3{\color{blue}2}\cdot \color{blue}{2^2}\cdot\left(x-\frac{1}{2}\right)^2=\\=-\frac 32\left(\color{blue}2\cdot x-\color{blue}2\cdot\frac{1}{2}\right)^2=-\frac 32(2x-1)^2$$
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