It is currently 27 Mar 2023, 12:14 |

Customized

for You

Track

Your Progress

Practice

Pays

- Mar
**27**### Should you waive the GMAT/GRE?

04:00 PM EDT

-05:00 PM EDT

Wondering if you should waive the GMAT/GRE? In this quick tips video, Personal MBA Coach Founder, Scott Edinburgh, discusses when it may be a good idea to waive the GMAT/GRE - Mar
**28**### MBA Research for Stellar MBA Applications | How to Select the Best Business Schools to Apply

08:30 AM PDT

-09:30 AM PDT

In this video, Jenifer Turtschanow, CEO of ARINGO Consulting, provides a complete framework to select the MBA programs to apply that fits to your profile and post MBA goals. The session will cover: Types of MBA Who are the top MBA programs, and more. - Mar
**28**### Decision Making Guide

04:00 PM PDT

-05:00 PM PDT

Wondering how to choose the right school and program type? Check out the Personal MBA Coach MBA Decision Making guide! - Mar
**29**### New hourly packages! The world has changed and so have we!

08:00 PM PDT

-09:00 PM PDT

Starting at $975 and offering up to $545 of savings, our new Amplify Hourly Packages offer structure and support in certain high-value areas (story development, resume, essays, interviewing) to help get your applications across the finish line. - Mar
**31**### Are you ready to take your career to the next level with an MBA?

09:30 AM EDT

-10:30 AM EDT

Join our Sia Admissions founder, Susan Berishaj, on March 31st to learn 3 steps for crafting admission committee-proof goals. Top-ranked business schools seek leaders with impactful and achievable goals. Let Susan help you make your dreams a reality. - Apr
**01**### Admissions to top DEFERRED MBA programs

06:00 PM PDT

-07:00 PM PDT

Start your DEFERRED MBA applications with us today! Our successful past admits to HBS, Wharton, Columbia and Haas deferred programs can attest to our expertise. Let our HBS 2+2 alum lead consultant and M7 school mentors help you secure your spot. - Apr
**03**### Get a headstart on the IIM One Year MBA admissions process.

06:00 PM IST

-07:00 PM IST

IIMReady is here to help experienced professionals gear up for the critical general awareness, sector knowledge, guesstimates and WAT component of the admissions process. Join us to get a headstart on the preparation.

x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.
[#permalink]
20 May 2020, 09:02

Expert Reply

8

Bookmarks

Question Stats:

x, y, and z are integers such that (\(x + 2)(y - 3)(z + 4) = 0\). Which of the following could be the value of xyz?

Indicate \(all\) such values.

❑ -9

❑ -5

❑ 0

❑ 6

❑ 7

❑ 12

❑ 27

_________________

Indicate \(all\) such values.

❑ -9

❑ -5

❑ 0

❑ 6

❑ 7

❑ 12

❑ 27

_________________

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.
[#permalink]
25 May 2020, 15:37

6

1

Bookmarks

This was a tricky one for me to see the solution, or even understand the solutions that were given!

The trick partially lies in recognizing that when determining the values of x, y, and z separately (before you even multiply them all together), you have to remember two things: 1) they're all integers and 2) at least ONE of their values has to generate a value of "0" as per the polynomial given. They don't ALL have to generate a value of 0, just one of them.

Stated differently: For the factors of x, y, and z that - when multiplied together - generate the number x*y*z, you can basically make them any integer as long as ONE of them solves the polynomial.

That means at least one of the following has to be true: x = -2, y = 3, or z = -4.

So when you're looking at the given answers "a" through "g" as a potential value of x*y*z, you know that the factors are all integers. That means that for a given answer value must have 2, 3, or 4 as a factor, otherwise the factors won't have the opportunity to multiply together to generate that answer value. (Remember: the other two-out-of-the-three values can become any integer, and you don't have to solve for them. And because the other two out of the three values can become any integer, you don't have to think about negatives and positives here because those hypothetical values have your back.)

Going through each of the answer options...

(a) -9 ---> 3 is a factor. YES, THIS IS A POSSIBLE ANSWER.

(b) -5 ---> neither 2 nor 3 nor 4 is a factor. NOT AN ANSWER.

(c) 0 ---> Take a step back on this one; you know that x or y or z could be set to 0 and you get 0 as an answer. YES, THIS IS A POSSIBLE ANSWER.

(d) 6 ---> 2 and 3 are factors. YES, THIS IS A POSSIBLE ANSWER.

(e) 7 ---> neither 2 nor 3 nor 4 is a factor. NOT AN ANSWER.

(f) 12 ---> 2 and 3 and 4 are all factors. YES, THIS IS A POSSIBLE ANSWER.

(g) 27 ---> 3 is a factor. YES THIS IS A POSSIBLE ANSWER.

A,C,D,F,G

The trick partially lies in recognizing that when determining the values of x, y, and z separately (before you even multiply them all together), you have to remember two things: 1) they're all integers and 2) at least ONE of their values has to generate a value of "0" as per the polynomial given. They don't ALL have to generate a value of 0, just one of them.

Stated differently: For the factors of x, y, and z that - when multiplied together - generate the number x*y*z, you can basically make them any integer as long as ONE of them solves the polynomial.

That means at least one of the following has to be true: x = -2, y = 3, or z = -4.

So when you're looking at the given answers "a" through "g" as a potential value of x*y*z, you know that the factors are all integers. That means that for a given answer value must have 2, 3, or 4 as a factor, otherwise the factors won't have the opportunity to multiply together to generate that answer value. (Remember: the other two-out-of-the-three values can become any integer, and you don't have to solve for them. And because the other two out of the three values can become any integer, you don't have to think about negatives and positives here because those hypothetical values have your back.)

Going through each of the answer options...

(a) -9 ---> 3 is a factor. YES, THIS IS A POSSIBLE ANSWER.

(b) -5 ---> neither 2 nor 3 nor 4 is a factor. NOT AN ANSWER.

(c) 0 ---> Take a step back on this one; you know that x or y or z could be set to 0 and you get 0 as an answer. YES, THIS IS A POSSIBLE ANSWER.

(d) 6 ---> 2 and 3 are factors. YES, THIS IS A POSSIBLE ANSWER.

(e) 7 ---> neither 2 nor 3 nor 4 is a factor. NOT AN ANSWER.

(f) 12 ---> 2 and 3 and 4 are all factors. YES, THIS IS A POSSIBLE ANSWER.

(g) 27 ---> 3 is a factor. YES THIS IS A POSSIBLE ANSWER.

A,C,D,F,G

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.
[#permalink]
20 May 2020, 09:02

Expert Reply

Post A Detailed Correct Solution For The Above Questions And Get A Kudos.

Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!

_________________

Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!

_________________

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.
[#permalink]
20 May 2020, 14:01

2

(x + 2)(y— 3)(z + 4) = 0

to make it zero; there are three possibilities: x = -2 or y = 3 or z = -4

if x = -2, only products 6, 0 and 12 can provide y and z as integers

if y = 3, only products -9, 0 and 27 can provide x and z as inegers

if z = -4, only product 12 and 0 can provide x and y as integers

Ans: 6, 0, 12, -9, 27

to make it zero; there are three possibilities: x = -2 or y = 3 or z = -4

if x = -2, only products 6, 0 and 12 can provide y and z as integers

if y = 3, only products -9, 0 and 27 can provide x and z as inegers

if z = -4, only product 12 and 0 can provide x and y as integers

Ans: 6, 0, 12, -9, 27

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.
[#permalink]
21 May 2020, 06:21

vndnjn wrote:

(x + 2)(y— 3)(z + 4) = 0

to make it zero; there are three possibilities: x = -2 or y = 3 or z = -4

if x = -2, only products 6, 0 and 12 can provide y and z as integers

if y = 3, only products -9, 0 and 27 can provide x and z as inegers

if z = -4, only product 12 and 0 can provide x and y as integers

Ans: 6, 0, 12, -9, 27

to make it zero; there are three possibilities: x = -2 or y = 3 or z = -4

if x = -2, only products 6, 0 and 12 can provide y and z as integers

if y = 3, only products -9, 0 and 27 can provide x and z as inegers

if z = -4, only product 12 and 0 can provide x and y as integers

Ans: 6, 0, 12, -9, 27

please provide a detailed solution

Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.
[#permalink]
21 May 2020, 10:55

2

Expert Reply

2

Bookmarks

The explanation above is already very efficient and simple.

We can use another alternative approach

Xyz we need the value which is multiplication

Now

x=-2

Y=3

Z=-4

So Xyz must be a multiple of 2,3, and 4. However, 4 it itself a multiple of 2 4=2^2

So the question boils down that the values possible for Xyz are a multiple of 2,3 or both which is 6

Among the answer choices A, C,D,F, and G are multiple of these factors

Hope now is more clear

_________________

We can use another alternative approach

Xyz we need the value which is multiplication

Now

x=-2

Y=3

Z=-4

So Xyz must be a multiple of 2,3, and 4. However, 4 it itself a multiple of 2 4=2^2

So the question boils down that the values possible for Xyz are a multiple of 2,3 or both which is 6

Among the answer choices A, C,D,F, and G are multiple of these factors

Hope now is more clear

_________________

Re: x, y, and z are integers such that (x + 2)(y 3)(z + 4) = 0.
[#permalink]
12 Dec 2022, 05:02

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

gmatclubot

Moderators:

Multiple-choice Questions — Select One or More Answer Choices |
||

## Hi Guest,Here are updates for you: |