Answer:
b = 259
Step-by-step explanation:
You want the smallest positive 'b' that makes x² +bx +2008 have integer solutions.
Factors
When the polynomial is factored, you have ...
(x +p)(x +q) = x² +(p+q)x +pq
Comparing this to the given expression, we see that ...
p + q = b
pq = 2008
This means the values of p and q will be divisors of 2008.
Factor pairs
The positive integer factor pairs of 2008 are ...
2008 = 1·2008 = 2·1004 = 4·502 = 8·251
The pair with the smallest sum is 8·251. The corresponding sum is ...
8 +251 = 259
The smallest positive value of 'b' that makes x² +bx +2008 have integer solutions is 259.
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Additional comment
The factorization is x² +259x +2008 = (x +8)(x +251).
