I know that if $x^2+bx+2008$ factors into a product of two binomials, then those binomials must have the form $(x+a)(x+c)$, where $a$ and $c$ are integers. Expanding this product, I get $x^2+(a+c)x+ac$. Comparing this with $x^2+bx+2008$, I see that $b=a+c$ and $2008=ac$. So I need to find two integers $a$ and $c$ that add up to $b$ and multiply to $2008$. One way to do this is to list the factors of $2008$ and see which pairs of them have a sum that is a positive integer. I can use prime factorization to find the factors of $2008$. I notice that $2008=2^3\cdot 251$. So the factors of $2008$ are $1, 2, 4, 8, 251, 502, 1004, 2008$. I can pair them up as follows: $(1, 2008), (2, 1004), (4, 502), (8, 251)$. I see that the pair $(8, 251)$ has the smallest sum, which is $259$. So $a=8$, $c=251$, and $b=a+c=259$. 
Find the smallest positive integer $b$ for which $x^2+bx+2008$ factors into a product of two binomials, each having integer coefficients.