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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.

User Neapolitan
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1 Answer

1 vote

Answer:

b = 259

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).

User David Soussan
by
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