Solve for Missing Denominator in 14x⁴-7x² = 7x² × Unknown

Let's examine the problem:

$\frac{14x^4-7x^2}{?}=7x^2$

First let's examine 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 possible common factor (meaning that the expression remaining in parentheses cannot be further factored by taking out a common factor):

$\frac{14x^4-7x^2}{?}=7x^2 \\ \downarrow\\ \frac{7x^2(2x^2-1)}{?}=7x^2 \\$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 on the right side as a fraction (using the fact that dividing a number by 1 does not change its value):

$\frac{7x^2(2x^2-1)}{?}=7x^2\\ \downarrow\\ \frac{7x^2(2x^2-1)}{?}=\frac{7x^2}{1}$

Let's continue solving the problem, thinking logically, and remember the fraction reduction operation, noting that in both the numerator and denominator and on both the right and left sides there exists the expression:$7x^2$, therefore we don't want to reduce from the numerator on the left side, however, the expression:

$2x^2-1$,

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:

$2x^2-1$

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

$\frac{7x^2(2x^2-1)}{?}=\frac{7x^2}{1} \\ \downarrow\\ \frac{7x^2(2x^2-1)}{\textcolor{red}{2x^2-1}}\stackrel{?}{= }\frac{7x^2}{1} \\ \downarrow\\ \boxed{\frac{7x^2}{1} \stackrel{!}{= }\frac{7x^2}{1} }$

Therefore choosing the expression:

$2x^2-1$

is indeed correct.

Which means that the correct answer is answer C.