Hi Mike…SAT 7, Section 4, Q6: I now see the shortcut here (that both sides of the equation are perfect squares,) but if I did expand and FOIL the left side, wouldn’t I still get the correct “a” values even though it takes longer? I can’t get it to work !! Can you please show the alternate path math steps? Or is recognizing the perfect squares the ONLY way to solve this one ? Thanks!

FOIL should work, sure. Plug in –3 for x and go to town:

    \begin{align*}(-3a+3)^2&=36\\9a^2-18a+9&=36\\a^2-2a+1&=4\\a^2-2a-3&=0\\(a-3)(a+1)&=0\end{align*}

That tells you that a could equal –1 or 3, and –1 is an answer choice.

You can also get this one by backsolving, though. That would actually be my first choice. You’re given x = –3. Try the answer choices for a until one works! Conveniently, I usually start with C when I backsolve, so I finish this one very quickly:

Choice C:

((-1)(-3)+3)^2=36

6^2=36

36=36 <– checks out