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Find the smallest positive integer \(b\) for which \(x^2+bx+2008\) factors into a product of two terms, each having integer coefficients.

 Jan 18, 2019
 #1
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The divisors of 2008  are

 

1, 2, 4, 8, 251, 502, 1004, 2008

 

"b"  will be smallest  when   we have  251 and 8

 

So

 

(x + 251) ( x + 8)  =   x^2 + 259x + 2008

 

So....the smallest value of "b"  =   259

 

 

cool cool cool

 Jan 18, 2019

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