Question 195·200 Super-Hard SAT Math Questions·Advanced Math

Choose $l=1$ and $m=2$, so $k=l+m=3$. Let $x_1=4$ be your test value of $x$. It avoids both excluded values, $0$ and $3$. Type those assignments, define $f$ as the original expression, and enter $f(x_1)$. Desmos shows $-2$.

Type $x_1+2k$. Desmos shows $10$, not $-2$, so $x+2k$ fails at an allowed input.

Type $\frac{x_1^2-k^2}{x_1-k}$. Desmos shows $7$, not $-2$. That form leaves out the second fraction being subtracted.

Type $\frac{x_1^2-2k^2}{x_1+k}$. Desmos shows approximately $-0.285714$, not $-2$. The original has $x-(l+m)=x-k$, so $x+k$ has the wrong sign.

Type $\frac{x_1^2-2k^2}{x_1-k}$. Desmos shows $-2$, matching the original. It's the only choice that matches at this allowed input. The algebra below shows why $\frac{x^2-2k^2}{x-k}$ works for every allowed input. Choice D.

The three terms involving $l$ and $m$ form a squared sum: $l^2+2lm+m^2=(l+m)^2=k^2$. The last equality uses the given $k=l+m$.

Factor $x$ out of the three subtracted terms in the top, and group $xl+xm$ in the bottom: $\frac{x^3-x(l^2+2lm+m^2)}{x^2-x(l+m)}-\frac{k^2}{x-(l+m)}$. Now the repeated $l+m$ is visible.

Use the squared sum above and $l+m=k$ in each denominator: $\frac{x^3-xk^2}{x^2-xk}-\frac{k^2}{x-k}$. Keep the minus sign in $x-k$.

Both the top and bottom of the first fraction have a whole factor of $x$. Factor it out: $\frac{x(x^2-k^2)}{x(x-k)}-\frac{k^2}{x-k}$.

The problem gives $x\ne0$, so dividing the first fraction's top and bottom by $x$ is allowed: $\frac{x^2-k^2}{x-k}-\frac{k^2}{x-k}$. Cancel whole factors, not pieces of a sum. The second fraction is still there.

The fractions now have the same denominator, so subtract their tops and keep $x-k$: $\frac{x^2-k^2}{x-k}-\frac{k^2}{x-k}=\frac{x^2-k^2-k^2}{x-k}=\frac{x^2-2k^2}{x-k}$. This is equivalent to the original wherever $x\ne0$ and $x\ne k$. Choice D.