+0  
 
0
60
3
avatar

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! :)

 Dec 23, 2020
 #1
avatar
0

The smallest b is 506: x^2 + 506x + 2008 = (x + 4)(x + 502)

 Dec 23, 2020
 #2
avatar
0

That is not correct.

Guest Dec 23, 2020
 #3
avatar+114325 
+1

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

 

 

cool cool cool

 Dec 23, 2020

12 Online Users

avatar
avatar