Question 177·Medium·Advanced Math

The denominator, the bottom of the fraction, can't be $0$. Since $x-2=0$ at $x=2$, the original expression requires $x\ne2$. Use $x=0$ for a quick test: it's allowed, and it makes the $x$ terms vanish.

Type $x_1=0$, then type $\dfrac{3x_1^2-12}{x_1-2}$. Desmos shows $6$, so any equivalent choice must also give $6$ at $x=0$.

Type each choice with $x_1$ in place of $x$. Desmos shows $-6$, $-6$, $4$, and $6$, in order. The first three fail at an allowed input; only $3x+6$ remains. One matching test wouldn't prove it on its own, but eliminating every other choice identifies the equivalent expression. Choice D.

Both terms on top share a factor of $3$. Factor it out of the whole numerator:

A difference of squares has the pattern $a^2-b^2=(a-b)(a+b)$. Here $x^2-4=x^2-2^2$, so its factors have opposite signs:

The entire factor $x-2$ multiplies the top and is the whole bottom. Cancel whole matching factors, not terms inside a sum. You can divide by $x-2$ only when it isn't zero, so keep the original restriction:

Distribute the $3$ across both terms:

So $3x+6$ equals the original fraction for every allowed $x$, even though the original is still undefined at $x=2$. Choice D.