GMAT Math : Expression simplification

One key technique for expression simplification GMAT Math questions is to figure out a ‘smart’ value to plug in.

Let’s start with an example:

What is the simplification of (x2 + 5x + 2)(y2 + 2x + 3)(z2 + 4x + 10)?

A. (60 + 214x + 58xy2 + 5xy2z2 + 19xz2 + 206x2 + 30x2y2 + x2y2z2 + 13x2z2 + 72x3 + 4x3y2 + 2x3z2 + 8x4 + 20y2 + 2y2z2 + 6z2)

B. (60 + 214x + 58xy2 + 5xy2z2 + 18xz2 + 207x2 + 30x2y2 + x2y2z2 + 13x2z2 + 72x3 + 4x3y2 + 2x3z2 + 8x4 + 20y2 + 2y2z2 + 6z2)

C. (60 + 214x + 58xy2 + 5xy2z2 + 40xz2 + 206x2 + 30x2y2 + x2y2z2 + 13x2z2 + 72x3 + 4x3y2 + 2x3z2 + 8x4 + 20y2 + 2y2z2 + 6z2)

D. (40 + 214x + 58xy2 + 5xy2z2 + 19xz2 + 206x2 + 30x2y2 + x2y2z2 + 13x2z2 + 72x3 + 4x3y2 + 2x3z2 + 8x4 + 20y2 + 2y2z2 + 6z2)

E. (13 + 214x + 58xy2 + 5xy2z2 + 19xz2 + 206x2 + 30x2y2 + x2y2z2 + 13x2z2 + 72x3 + 4x3y2 + 2x3z2 + 8x4 + 20y2 + 2y2z2 + 6z2)

Readers are encouraged to give the above problem a shot. Challenge yourself to not spend more than 45 sec – 1 min 🙂

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There is absolutely no way we are multiplying and simplifying the original equation (x2 + 5x + 2)(y2 + 2x + 3)(z2 + 4x + 10) as it would take a ton of time.

What we want to do is to exclude options by plugging real numbers. What sort of numbers do we plug in? We want to use the easiest numbers like 0 and 1 which when raised to a power remain the same and do not need additional calculation.

We start with 0

(x2 + 5x + 2)(y2 + 2x + 3)(z2 + 4x + 10) when x, y and z = 0 becomes

2 x 3 x 10 = 60

Plugging into the answer options

Option (D) and (E) are out since they yield 40 and 13 when x, y and z are plugged with 0

It would be great if we could eliminate all but 1 option by plugging 0. Q50+ questions are seldom this easy but we are ready for the challenge 🙂

Let’s go with 1 now.

(x2 + 5x + 2)(y2 + 2x + 3)(z2 + 4x + 10) when x, y and z = 1 becomes

8 x 6 x 15 = 720

Plugging into the answer options (A), (B) and (C)

Only (A) and (B) satisfy this.

Now, we are left with (A) and (B) and 0 and 1s are unable to differentiate them

A powerful tool to tackle such as situation is to find the ‘algebraic difference’ between (A) and (B)

Which is essentially (A) – (B) = (60 + 214x + 58xy2 + 5xy2z2 + 19xz2 + 206x2 + 30x2y2 + x2y2z2 + 13x2z2 + 72x3 + 4x3y2 + 2x3z2 + 8x4 + 20y2 + 2y2z2 + 6z2)  – (60 + 214x + 58xy2 + 5xy2z2 + 18xz2 + 207x2 + 30x2y2 + x2y2z2 + 13x2z2 + 72x3 + 4x3y2 + 2x3z2 + 8x4 + 20y2 + 2y2z2 + 6z2)

= xz2 – x2 (Notice the subtraction is straightforward as most of the terms are the same – this is by design as the GMAT want to make it hard to differentiate between the two options by making all terms almost the same – we leverage this to our benefit)

The next steps is to figure out what simple value would make the difference ‘non zero’

Plugging xz2 – x2 with 0 leads to 0

But plugging with x = 1, y = 0 z = 0 yields -1

The set of values which lead to a non-zero differential is what we use as the next set of values to test with.

x = 1, z = 0 and y = 0

(x2 + 5x + 2)(y2 + 2x + 3)(z2 + 4x + 10)

(8) x (5) x (14) = 560

Plugging into option (A) we get 560 and into B we get 561 (in fact, we didn’t have to plug into B – as soon as 1 option works out we are assured the other option will give a different value because of the differential)

We go with option (A)

A recap of the algorithm is below