Problem 14 Polynomials
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Problem 14: Polynomials Problem: For how many integers x is
x3 +2x2 +3x+4 x+5
Art of Problem Solving
an integer?
Solution: We can write x3 + 2x2 + 3x + 4 = P (x) · (x + 5) + Q, for some integer Q and some integer polynomial P (x). Plugging in x = −5 into this form gives P (−5) · (−5 + 5) + Q = Q = −86. Thus,
x3 +2x2 +3x+4 x+5
=
P (x)·(x+5)−86 x+5
= P (x) −
86 x+5
Clearly, P (x) is an integer, so we need (x + 5)|86. 86 has 8 factors: ±1, ±2, ±43, ±86, so there are 8 possible integer values for x + 5, and thus 8 possible values for x.
Solution was written by The Zuton Force and compiled from Art of Problem Solving Forums.
x3 +2x2 +3x+4 x+5
Art of Problem Solving
an integer?
Solution: We can write x3 + 2x2 + 3x + 4 = P (x) · (x + 5) + Q, for some integer Q and some integer polynomial P (x). Plugging in x = −5 into this form gives P (−5) · (−5 + 5) + Q = Q = −86. Thus,
x3 +2x2 +3x+4 x+5
=
P (x)·(x+5)−86 x+5
= P (x) −
86 x+5
Clearly, P (x) is an integer, so we need (x + 5)|86. 86 has 8 factors: ±1, ±2, ±43, ±86, so there are 8 possible integer values for x + 5, and thus 8 possible values for x.
Solution was written by The Zuton Force and compiled from Art of Problem Solving Forums.
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