Solve for Missing Denominator in (6x³-3x²)/(?) = 3x²

Examine the following problem:

$\frac{6x^3-3x^2}{?}=3x^2$

Note that in the numerator of the fraction on the left side 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):

$\frac{6x^3-3x^2}{?}=3x^2 \\ \downarrow\\ \frac{3x^2(2x-1)}{?}=3x^2 \\$In factoring, we used of course the law of exponents:

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

Proceed to express the expression on the right side as a fraction ( by using the fact that dividing a number by 1 does not change its value):

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

Continue solving the problem and remember the fraction reduction operation. Note that in both the numerator on the right side and on the left side there is the expression:$3x^2$, hence we don't want to reduce from the numerator on the left side. However, the expression:

$2x-1$,

is not found in the numerator on the right side (which is the fraction after reduction) Therefore we can conclude that this expression needs to be reduced from the numerator on the left side, so the missing expression must be none other than:

$2x-1$

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

$\frac{3x^2(2x-1)}{?}=3x^2 \\ \downarrow\\ \frac{3x^2(2x-1)}{\textcolor{red}{2x-1}}\stackrel{?}{= }\frac{3x^2}{1} \\ \downarrow\\ \boxed{\frac{3x^2}{1} \stackrel{!}{= }\frac{3x^2}{1} }$

Therefore choosing the expression:

$2x-1$

is indeed correct.

Which means that the correct answer is answer B.