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