SECTION B: DATA INTERPRETATION AND QUANTITATIVE ABILITY

54. F(x) is a fourth order polynomial with integer coefficients and with no common factor. The roots of F(x) are –2, –1, 1, 2. If p is a 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

Solution:

F(p) = (p – 2) (p – 1) (p + 1) (p + 2).

Multiplying F(p) with p, we get p.F (p) = (p – 2) (p – 1) p(p + 1) (p + 2).

Which is the product of 5 consecutive integers, so it is divisible by 5! or 120.

Since p is a prime number, we can conclude that F(p) must be divisible by 120 because p and 120 will be coprimes.

In p.F(p) = (p – 2) (p – 1)p (p + 1) (p + 2) the expression (p – 2) (p – 1)p, which are 3 consecutive integers, must be divisible by 3! or by 6.

Again p and 6 will be coprimes since p is a prime number.

So (p – 2) (p – 1) is divisible by 6 .

Similarly we can conclude that (p + 1) (p + 2) will also be divisible by 6

So (p – 2) (p – 1). (p + 1). (p + 2) = F(p) is divisible by 6 ´ 6 i.e. 36.

Thus F(p) is divisible by 120 and 36 and hence by LCM of (36, 120). i.e. by 360.

Now to show that the maximum common factor for F(p) is 360, we need to check for some values of p for which the only common factor is 360.

For p = 101 and for p = 103, we get

F(101) = 360. 3.5.11.17.103. and

F (103) = 360.2.7.13.17.101.

In the above two cases we get the common factor of F(p)as 360.17. But we can always show that 17 will not be a common factor for all values of p.

For example, when p = 107 or 109 or 113 we will not have 17 as a factor of F(p).

Hence we can conclude that 360 is the largest integer which divides F(p) for any value of p.

We can see, other than 360 there are no common factors so we can conclude that the largest integer which divides F(p) for all values of p is 360. Choice (D)