We all know not to divide by zero. It is a rule from middle school—if not earlier—and the reasons for it are pretty straightforward.
If you look at the graph of y = 1/x, the y value approaches +âˆž as x approaches zero from the right, and the y value approaches —âˆž as x approaches zero from the left. But the graph never reaches x = 0, because you cannot divide by zero. Dividing 1 by smaller and smaller fractions results in larger and larger quotients, because many tiny bits can fit into one whole. But you can’t answer the question of how many zeros fit into 1; the question doesn’t make sense conceptually.
All this is interesting, and the history of zero is at least a little bit interesting, too. But for the purposes of the GMAT, we have already thought much more about zero than we have to. If we remember not to divide by zero, we have remembered everything we need to know for test day. Or have we?
Here is a problem where aspiring GMAT 800′s tend to forget that dividing by zero can cause trouble on the Quant section:
If (x + 4)(3x + 1) = 3x2 + x, what is a possible value of x?
There are a couple potential approaches to this problem. We could FOIL the expression on the left side of the equation. That won’t take too long, but if we’re really up on our game, we might notice that if we factor an x from the expression on the right side of the equation, there will be a (3x + 1) on both sides, which will let us cancel and simplify. That would be faster, and every second helps, so let’s use that method.
(x + 4)(3x + 1) = 3x2 + x â†’
(x + 4)(3x + 1) = x(3x + 1)
We cancel the (3x + 1) on both sides, giving us:
x + 4 = x
Now we subtract x from both sides and get:
4 = 0
Wait a minute. Something went wrong. It is quite certain that 4 does not equal 0, so what happened? We can go over our calculations, but we didn’t make any errors. And this is the GMAT; 75 minutes are ticking away fast, so we don’t have time to ponder the rift in the universe that allows 4 = 0. Let’s just do the problem again really quickly with the first method. (We’ll take a second look after we finish.)
(x + 4)(3x + 1) = 3x2 + x â†’
3x2 + x + 12x + 4 = 3x2 + x
Now we combine like terms by subtracting 3x2 andÂ x from both sides, and we solve:
12x + 4 = 0
12x = —4
x = —1/3
So, —1/3 is a possible value of x, and answer choice D is correct.
Now let’s take a look at what went wrong when we factored and canceled. When we first pulled out an x, giving us (x + 4)(3x + 1) = x(3x + 1), everything was going fine. We hadn’t broken any rules yet.
And then we canceled the (3x + 1). When we cancel in this situation, what we are doing is dividing both sides by (3x + 1). The factors essentially go away, since (3x + 1)/(3x + 1) is always equal to 1. Except when (3x + 1) is equal to zero! Hindsight is 20/20, so let’s plug in x = —1/3, and sure enough it turns out that (3x + 1) is zero.
The takeaway:Â We can never divide by a variable, or by a variable expression, unless we know that the variable or expression does not equal zero. Remember, canceling is dividing, too.
Keep this in mind and you’ll avoid the head-scratching realization that 4=0. This will save you some troubling philosophical pondering, not to mention a lot of valuable time on test day.