F(x)
is a fourth order polynomial with integer coefficients and no common factor.
The roots of F(x) are -2,-1,1,2. If p is prime number greater than 97, then the
largest integer that divides F(p) for all values of p is:

A.72 B. 120 C. 240 D. 360 E. None of the above

A.72 B. 120 C. 240 D. 360 E. None of the above

Solution follows
here:

**Solution:**
The roots of F(x)
are -2,-1,1,2 => F(x) = (x+2)(x+1)(x-1)(x-2)

F(p) = (p+2)(p+1)(p-1)(p-2)
= (p+2)(p+1)(p)(p-1)(p-2)/p

=> p*F(p) = (p+2)(p+1)(p-1)(p-2)
= (p+2)(p+1)(p)(p-1)(p-2)

p*F(p)
is a product of

=> it is

**5 consecutive numbers**=> it is

**divisible by 5**---(1)
Middle
number p is prime => obviously it is odd => (p-1) and (p+1) are
consecutive even numbers => For two consecutive even numbers, one must be a multiple
of 4 and the other must be a multiple of 2 => p*F(p) is

**divisible by 4*2**ie.,**8**---(2)
One of the

**three consecutive numbers**(p+2),(p+1),p is a multiple of 3 => as p is prime, either (p+2) or (p+1) is a multiple of 3 => similarly, either, (p-2) or (p-1) is a multiple of 3 => p*F(p) is**divisible by 3*3**ie.,**9**---(3)
From (1),(2) and (3), p*F(p) is

**divisible by 5*8*9**ie.,**360****=>**As p is prime,

**F(p) is**

**divisible by 360**

__Note:__
As it is given that p > 97, it should not be one of 2 or 3 or 5 and hence we can surely say that, "As p*F(p) is divisible by 360

**,**F(p) is divisible by 360". If not, p can be one of 2 or 3 or 5, then F(p) can be divisible by a 360/2 or 360/3 or 360/5 ie., 180 or 120 or 72.**Answer (D)**
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