Question 63·Hard·Advanced Math

The numbers $-6$ and $1$ multiply to $-6$ and add to $-5$, so factor the first denominator:

The domain is the set of inputs where an expression is defined. The original denominators rule out $x=6$, $x=-1$, and $x=-2$. Don't use those inputs to test a choice.

Type the original as $f(x)$, then type $f(0)$. Desmos prints $4$. The input $0$ is allowed, so any equivalent choice must also give $4$ there.

Type each of the other three choices with $0$ in place of $x$. Desmos prints $-4$ for each. None can be equivalent to the original, which gives $4$ at $x=0$.

Show the graph and type the remaining complete choice as $y=$ followed by that choice. Its curve covers $f(x)$ across every allowed interval, not only at $0$. The equivalent expression is $\dfrac{(x-2)(x^{2}+22x+24)}{(x-6)(x+1)(x+2)}$. Choice D.

Factor out $3$:

The difference of squares rule turns a square minus a square into two factors:

Keep both factors for now; a factor in a numerator doesn't change what denominator the fraction needs.

Both terms share $2$, so factor it out:

Now $x-2$ appears in both numerators, which will help after the denominators match.

The least common denominator uses every factor from the two denominators: $(x-6)(x+1)(x+2)$. The first fraction's denominator is missing $x+2$, so multiply its top and bottom by $x+2$:

Having $x+2$ on top doesn't mean it's already in the denominator.

The second denominator is missing $(x-6)(x+1)$. Multiply both its top and bottom by that product:

The denominators now match, so subtract the entire second numerator:

Keep the minus sign in front of that whole product; dropping it would change the expression.

Both terms on top contain $x-2$, so use factoring, or distribution backward:

Only whole factors can be pulled out this way.

Square the first factor:

Multiply the second pair:

Substitute those results:

Distribute, including the negative $2$:

Combine terms with the same power of $x$:

Replace the bracket with $x^{2}+22x+24$. The result is $\dfrac{(x-2)(x^{2}+22x+24)}{(x-6)(x+1)(x+2)}$, with the original exclusions $x\ne6,-1,-2$ still in force. Choice D.