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 \(f(x)\)