Bunuel wrote:

\(f(x)=(4x+1)(4x–3)\) for all positive integers x. Which of the following cannot be a factor of f(x)?

I. 7

II. 16

III. 25

A. I only

B. II only

C. I and II only

D. I and III only

E. I, II, and III

Working through numbers has already been explained so here's an alternate approach

\(4x+1\) & \(4x-3\) are ODD as \(x\) is an integer. Product of Odd numbers will not be divisible by Even number. So \(16\) cannot be a factor of \(f(x)\)

So options A & D are out. Now I is present in options C & E, if we can eliminate I then the answer will be B.

if \(7\) is a factor of \(f(x)\), then either \(4x+1\) or \(4x-3\) has to be a multiple of \(7\)

so if, \(4x-1=7k => x=\frac{7k+1}{4}\), where \(x\) has to be an integer. clearly if \(k=1\) then \(x=2\), an integer

or \(4x-3=7k=> x=\frac{7k+3}{4}\), if \(k=3\) then \(x=6\), an integer. So we can conclude \(7\) can be a factor of \(f(x)\)

So we have eliminated Option A, C, D & E

Answer: Option

B---------------------------------------------------------

just to test \(25\) as a factor of \(f(x)\):

Now if \(4x+1=25 => x=6\) an integer so \(25\) can be a factor of \(4x+1\). Similarly if \(4x-3=25=>x=7\). hence \(25\) can also be a factor of \(4x-3\). so we know \(25\) can be a factor of \(f(x)\)