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

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For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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New post 12 Apr 2019, 10:07
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Question Stats:

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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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For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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New post 12 Apr 2019, 10:26
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
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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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New post 13 Apr 2019, 07:15
Hi can anyone offer an explanation of the question?
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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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New post 13 Apr 2019, 07:18
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
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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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New post 13 Apr 2019, 07:40
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.
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For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by...  [#permalink]

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New post 13 Apr 2019, 07:44
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.
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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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New post 13 Apr 2019, 07:49
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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.
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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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New post 13 Apr 2019, 22:03
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
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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] 13 Apr 2019, 22:03
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