The equation is true for all valuesof , where a is a constant. What is the value of a?A) – 4
B) – 2
C) 4
D) 16
This type of problem is not that hard if you recognize what’s required. But
when you’re under pressure—particularly time pressure—you can really lose
focus. If someone were to take a “brain transcript” of you trying to do this
problem, here’s what it might look like:
“What is THIS monstrosity? Wait . . . ooooh-kay. It looks like one of those
rational functions that I have to divide. Yeah, poly . . . Polly . . . no-mials. Yeah.
What the hell does Polly know? Probably more than me. . . . All right, all right,
what am I looking at? I don’t know a, so how can I divide by the denominator?
Crap. Well, what’s this garbage on the right side? I can never remember what the
result of division is called. Dividends? Divisors? Shoot! I’m losing time. What
do I have left? Two minutes? If I don’t get this right, I’ll probably lose like 20
points. Maybe I should guess. But I have no frickin’ idea. Hmm . . . What about
if I divide? In that case, the 12x^2 part is going to become –3x after it’s divided by
the “ax.” I wish that idiot in front of me would stop tapping his foot. Tap. Tap.
Tap. Tap. There he goes again. What’s up with his hair? Tap. Tap. He looks like
he just came from the gym. Gross! Hey, guy, you know, gyms have these great
things called “showers.” Okay, okay, okay. Back to this problem. If 12x^2 is –3x,
then 12/a is going to be –3. What does a have to be, in order to get –3 as the first
term? Well, 12/a = –3. So I can multiply both sides by a to cancel, and then
divide both sides by –3. That gives me –4. So I don’t need to care about ANY of
the rest of this, the remainder and all that, because a is just –4. Bam, the
answer’s A. Fill in A.”
Well, you got the answer right, but only after much wasted thought. The
worst mistake was checking the time. Do this only between problems, never in
the middle of a problem.