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Manager  G
Joined: 21 Feb 2019
Posts: 125
Location: Italy
For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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7 00:00

Difficulty:   95% (hard)

Question Stats: 40% (02:30) correct 60% (02:11) wrong based on 58 sessions

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For how many values of $$a$$ is $$( x - 1)(x^2 - a^2)(x^2 - a - 1)$$ divisible by $$x^2 + x - 2$$?

A. 1
B. 2
C. 3
D. 5
E. None

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Intern  B
Joined: 07 Mar 2019
Posts: 29
For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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1
1
SOLUTION:

x^2 + x - 2 -> (x-1)(x+2)

(x-1)(x^2-a^2)(x^2-a-1) / (x-1)(x+2) -> (x^2-a^2)(x^2-a-1) / (x+2) -> (x-a)(x+a)(x^2-a-1) / (x+2) [we must find values of a that yield (x+2)]

We can first find two values with (x-a) & (x+a): a = 2 or a = -2

For (x^2-a-1), this is difficult, but can be done.

EXAMPLE:

If a is a value (let's say 2), then (x^2-a-1) would resolve to the form of (x^2-3), and "a" would represent sqrt(a+1): (x-sqrt(3))(x+sqrt(3))

Thus, we need to satisfy the following equation: x+2 = x + sqrt(a+1) -> 2 = sqrt(a+1) -> a + 1 = 4 -> a = 3

Thus, there are three possible values of a: -2, 2 & 3
Manager  B
Joined: 17 Sep 2017
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Re: For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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Hi can anyone offer an explanation of the question?
Intern  B
Joined: 07 Mar 2019
Posts: 29
Re: For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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abhishek31 wrote:
Hi can anyone offer an explanation of the question?
I posted the solution above you. What’s not clear?

Posted from my mobile device
Manager  B
Joined: 17 Sep 2017
Posts: 101
Re: For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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georgethomps wrote:
abhishek31 wrote:
Hi can anyone offer an explanation of the question?
I posted the solution above you. What’s not clear?

Posted from my mobile device

Hi, In the solutions provided you above, the part mentioned below

"Thus, we need to satisfy the following equation: x+2 = x + sqrt(a+1) -> 2 = sqrt(a+1) -> a + 1 = 4 -> a = 3"

Shouldn't have 2 solutions 1) x +sqrt(a+1) & 2nd x - sq(a+1) ?

In total taking all the possible solutions to 4.
Intern  B
Joined: 07 Mar 2019
Posts: 29
For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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abhishek31 wrote:
georgethomps wrote:
abhishek31 wrote:
Hi can anyone offer an explanation of the question?
I posted the solution above you. What’s not clear?

Posted from my mobile device

Hi, In the solutions provided you above, the part mentioned below

"Thus, we need to satisfy the following equation: x+2 = x + sqrt(a+1) -> 2 = sqrt(a+1) -> a + 1 = 4 -> a = 3"

Shouldn't have 2 solutions 1) x +sqrt(a+1) & 2nd x - sq(a+1) ?

In total taking all the possible solutions to 4.
Not quite. Although taking a square root can yield a positive or negative number, the number inside sqrt() still hs to equal 4. This can’t be done with two values of “a”, which is what the question is asking for.
Intern  B
Joined: 26 Jan 2018
Posts: 5
Location: United Arab Emirates
GMAT 1: 720 Q48 V41 GPA: 4
Re: For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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1
Ans C

x2-a-1 can be factorized as (x-1)(x+2).

Now (x-1) cancels out (x-1) in the numerator.
we are left with (x-a)(x+a)(x2-a-1) to be divided by (x+2).

only values that will give us (x+2) in the numerator are 2, -2 and 3.
Intern  B
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Posts: 12
Concentration: Organizational Behavior, Strategy
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Re: For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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1
lucajava wrote:
For how many values of $$a$$ is $$( x - 1)(x^2 - a^2)(x^2 - a - 1)$$ divisible by $$x^2 + x - 2$$?

A. 1
B. 2
C. 3
D. 5
E. None

Let's factor the divisor: $$x^2 + x - 2$$ into $$(x-1)(x+2)$$

Now we see we can cancel the $$(x-1)$$ terms from both equations, leaving us with:
Dividend: $$(x^2 - a^2)(x^2 - a - 1)$$, Divisor: $$(x+2)$$

The task now is to make $$(x+2)$$ appear by altering the value of $$a$$
We can factor the dividend further:
$$(x - a)(x+a)(x^2 - a - 1)$$

Now, let's see if we can tackle each part:

If $$a = 2, (x+a) =$$ (x+2)
If $$a = -2, (x-a) =$$ (x+2)
If $$a = 3, (x^2 - a - 1) = (x^2 - 3 -1) = (x^2-4)$$ factored (x+2)(x-2)

The three possible solutions: 2,-2,3
Manager  G
Joined: 08 Apr 2019
Posts: 148
Location: India
GPA: 4
Re: For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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tinklepoo wrote:
lucajava wrote:
For how many values of $$a$$ is $$( x - 1)(x^2 - a^2)(x^2 - a - 1)$$ divisible by $$x^2 + x - 2$$?

A. 1
B. 2
C. 3
D. 5
E. None

Let's factor the divisor: $$x^2 + x - 2$$ into $$(x-1)(x+2)$$

Now we see we can cancel the $$(x-1)$$ terms from both equations, leaving us with:
Dividend: $$(x^2 - a^2)(x^2 - a - 1)$$, Divisor: $$(x+2)$$

The task now is to make $$(x+2)$$ appear by altering the value of $$a$$
We can factor the dividend further:
$$(x - a)(x+a)(x^2 - a - 1)$$

Now, let's see if we can tackle each part:

If $$a = 2, (x+a) =$$ (x+2)
If $$a = -2, (x-a) =$$ (x+2)
If $$a = 3, (x^2 - a - 1) = (x^2 - 3 -1) = (x^2-4)$$ factored (x+2)(x-2)

The three possible solutions: 2,-2,3

I believe none of the above solutions caters to the question that many would have, as to how come a = 3.

I followed the same procedure for getting a = 2, -2

Now, for x^2 - a - 1, for it to be divisible by x+2, I went by the belief that it has to be in the form (x+2)(x-2), i.e. x^2 - 4
Equating both against each other, x^2 - a - 1 = x^2 - 4

x^2 - (a + 1) = x^2 - 4, and HENCE, a+1 = 4, and a = 3

So, with +2, -2, and +3, we have 3 solutions, in total Re: For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...   [#permalink] 29 May 2019, 19:48
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