An even positive integer 'x' has 'y' positive integral facto

Probably the easiest way would be to try different numbers:
If \(x=2\) then \(y=2\) --> \(4x=8=2^3\) and \(# \ of \ factors=4=2y\);
If \(x=2^2\) then \(y=3\) --> \(4x=16=2^4\) and \(# \ of \ factors=5=\frac{5y}{3}\);
Two different answers for two values of \(x\), hence we cannot determine the # of factors of \(4x\).

Answer: E.


THEORY:

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
For more on number properties check: math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION:

Given: \(x\) is even -->so \(x=2^p*b^q\), where \(b\) is some other prime factor of \(x\) (other than 2) and \(q\) is its power (note that x may or may not have other primes, this is just an example). The number of all factors of \(x\) is \(y=(p+1)(q+1)\) so:
if \(p=1\) then \(y=2(q+1)\);
if \(p=2\) then \(y=3(q+1)\);
if \(p=3\) then \(y=4(q+1)\);
....

Now, \(4x=2^2*x=2^{p+2}*b^q\) and \(4x\) will have \((p+2+1)(q+1)=(p+3)(q+1)\), so:
if \(p=1\) then \(# \ of \ factors=4(q+1)=2y\);
if \(p=2\) then \(# \ of \ factors=5(q+1)=\frac{5y}{3}\);
if \(p=3\) then \(# \ of \ factors=6(q+1)=\frac{6y}{4}\);
....

So # of factors of \(4x\) depends on the initial power of 2 in \(x\).

Answer: E.