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

 Feb 28, 2021
 #1
avatar+283 
+1

The factors of 2008 in pairs are \((1, 2008), (2, 1004), (4, 502), (8, 251)\). The sum of the factors closest to each other will yield the smallest possible value for b, so the answer is \(8+251=\boxed{259}\)

The factored quadratic would look like this: \((x+8)(x+251)\)

 Feb 28, 2021
 #2
avatar+112861 
0

jxt516:

 

Why don't you respond to my question? 

 

It is rude to not respond.

 

https://web2.0calc.com/questions/help-please_88636

 Mar 1, 2021
edited by Melody  Mar 1, 2021
 #3
avatar+112861 
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jxc, please explain to us how Textot got his answer.   Explain the reasoning behind it.

 

If you do not know then ask Textot questions about it.

Encourage or even push him/her to guide you through it.

 Mar 1, 2021
edited by Melody  Mar 1, 2021
 #4
avatar+201 
+1

I think you factor 2008, and using FOIL you find that (8,251) is the smallest

 

I might be wrong though

jxc516  Mar 1, 2021
 #5
avatar+112861 
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Since you do not understand you will need to ask Textot to explain.

Melody  Mar 1, 2021
 #6
avatar+201 
+3

Oh wait I think I understand

 

When you use foil to expand, you multiply the variables together, then the sum of the products of the inside and outside terms, and the constants. Example: (x+2)(x-3)=x^2-x-6. 

 

Because of this, we need to factorize 2008, so for the middle term we get  x+2008x=2009x, 2x+1004x=1006x, 4x+502x=506x, 8x+251x=259x, and we take the smallest coefficient and that's our answer

jxc516  Mar 1, 2021
 #7
avatar+112861 
+2

I think you have it!  Good work!

 

For any quadratic  \(ax^2+bx+c\)

 

\(\text{if the roots are } \alpha \text{ and } \beta \)

 

then

 

\(\alpha+\beta = \frac{-b}{a}\)

 

and

 

\(\alpha\beta = \frac{c}{a}\)

 

 

So in this case the roots multiply to 2008/1 =2008

Melody  Mar 1, 2021

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