Find the smallest positive integer b for which x^2+bx+2008 factors into a product of two binomials, each having integer coefficients.
Thank you so much! :)
The smallest b is 506: x^2 + 506x + 2008 = (x + 4)(x + 502)
That is not correct.
Factors of 2008 =
1 2 4 8 251 502 1004 2008
The smallest value for b is
( x + 8) ( x +251) = x^2 + ( 251 + 8) x + 2008 = 259
