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For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by 12 Aug 2022, 01:30
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C
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45% (02:18) correct 55% (02:11) wrong based on 109 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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C
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For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by 12 Aug 2022, 08:31
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factoring the term x^2+x-2 as (x+2)(X-1), we see that for the term (x-1)(x^2-a^2)(x^2-a-1), x-1 will cancel out leaving only (x^2-a^2)(x^2-a-1), for a=2, X^2-a^2 factorizes to (x-2)(x+2). Similarly for a=-3, X^2-a-1 = (X^2-2^2).
Hence a=2,-3 two values are possible
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For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by 17 Aug 2022, 00:31
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1. x^2+x-2 = (x+2)(X-1)
2. (x-1)(x^2-a^2)(x^2-a-1) = (x-1)(x-a)(x+a)(x^2-a-1)

If “2” is divisible by “1”, then “1” is factor of “2”. Which means (x-a) or (x+a) = (x+2), which means a=2 or -2.
Or (x+2) is a factor of (x^2-a-1). Consider there is “x^2” and no “nx” where n is a constant number. then if (x^2-a-1) can be (x^2-m^2) where m is a constant number. Then (x^2-a-1)=(x^2-m^2)=(x+2)(x-2), then -a-1=-4, a =3

In conclusion, a can be 2, -2 or 3

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For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by 17 Aug 2022, 01:32
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maybe 3.
logic is this:
in which cases "a" makes part of multiplication equal to 0.
hence, a=x, a=-x, a=x^2-1
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For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by 17 Aug 2022, 04:25
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\(x^2 + x - 2 = (x+2)(x-1)\)

\((x - 1)(x^2 - a^2)(x^2 - a - 1)\)

Now, all values of \(a\) which when factorised with the formula \(x^2 - a^2\) should result in either \((x+2)\) as we already have a \((x-1)\) in the numerator

Let us consider the following values of \(a\):

Case 1 (\(a = 2\)):
\((x - 1)(x^2 - a^2)(x^2 - a - 1) \\
= (x - 1)(x^2 - 2^2)(x^2 - 3) \\
= (x - 1)(x - 2)(x + 2)(x^2 - 3)\) is divisible by \((x + 2)(x - 1)\)

Case 2 (\(a = 3\)):
\((x - 1)(x^2 - a^2)(x^2 - a - 1) \\
= (x - 1)(x^2 - 3^2)(x^2 - 4) \\
= (x - 1)(x - 3)(x + 3)(x - 2)(x+2)\) is divisible by \((x + 2)(x - 1)\)

Case 3 (\(a = -2\)):
\((x - 1)(x^2 - a^2)(x^2 - a - 1) \\
= (x - 1)(x^2 - (-2)^2)(x^2 + 1) \\
= (x - 1)(x - 2)(x + 2)(x^2 + 1)\) is divisible by \((x + 2)(x - 1)\)

Total 3 values of a (\(2,-2,3\))

Answer - C

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For how many values of a (x-1)(x^2-a^2)(x^2-a-1) is divisible by 29 Aug 2024, 14:42
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