Solve for the Denominator in (4x³-2x)/? = 2x

Let's examine the problem:

$\frac{4x^3-2x}{?}=2x$

First let's check that in the numerator of the left fraction there is an expression that can be factored using factoring out a common factor, we will therefore factor out the largest common factor possible (meaning that the expression in parentheses cannot be further factored by taking out a common factor):

$\frac{4x^3-2x}{?}=2x \\ \downarrow\\ \frac{2x(2x^2-1)}{?}=2x$

In factoring, we used of course the law of exponents:

$\bm{a^{m+n}=a^m\cdot a^n}$

Now let's write the expression in the right side as a fraction (using the fact that dividing a number by 1 does not change its value):

$\frac{2x(2x^2-1)}{?}=2x \\ \downarrow\\ \frac{2x(2x^2-1)}{?}=\frac{2x}{1}$

Let's continue solving the problem, thinking logically, and remember the fraction reduction operation, note that in both the numerator and denominator of the left fraction and right fraction there exists the expression:$2x$, therefore we don't want to reduce from the numerator of the left fraction, however, the expression:

$2x^2-1$,

is not found in the numerator of the right fraction (which is the fraction after reduction) therefore we can conclude that this expression needs to be reduced from the numerator of the left fraction, so the missing expression must be:

$2x^2-1$

Let's verify that with this choice we indeed get the expression on the right side: (reduction sign)

$\frac{2x(2x^2-1)}{?}=\frac{2x}{1} \\ \downarrow\\ \frac{2x(2x^2-1)}{\textcolor{red}{2x^2-1}}\stackrel{?}{= }\frac{2x}{1} \\ \downarrow\\ \boxed{\frac{2x}{1} \stackrel{!}{= }\frac{2x}{1} }$

Therefore choosing the expression:

$2x^2-1$

is indeed correct.

Which means that the correct answer is answer A.