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x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.

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

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Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink] New post 20 May 2020, 09:02
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Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
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Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink] New post 20 May 2020, 14:01
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(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
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Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink] New post 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

please provide a detailed solution
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Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink] New post 21 May 2020, 10:55
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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
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Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0. [#permalink] New post 25 May 2020, 15:37
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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 :)
Re: x, y, and z are integers such that (x + 2)(y— 3)(z + 4) = 0.   [#permalink] 25 May 2020, 15:37
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