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# help help help

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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

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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
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jxt516:

Why don't you respond to my question?

It is rude to not respond.

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

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

Mar 1, 2021
edited by Melody  Mar 1, 2021
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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
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Since you do not understand you will need to ask Textot to explain.

Melody  Mar 1, 2021
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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
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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