Factorial of Quadratic Expression

Find the sum of all values of x x for the equation: ( x 2 + 5 x + 6 ) ! = 1 (x^2+5x+6)!=1

Clarification : The factorial in this problem is defined for integers 0 \geq 0 (i.e. the Gamma function is not considered).


The answer is -10.

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2 solutions

Alex G
Apr 18, 2016

Solving y ! = 1 y!=1

The solutions y = 1 y=1 and y = 0 y=0 are appearent by inspection. It is easily seen that there are no other positive solutions, as the factotrial increases uniformly over the positive integers.

Using only the integral solutions, we have x 2 + 5 x + 6 = 1 x^2+5x+6=1 and x 2 + 5 x + 6 = 0 x^2+5x+6=0 . By Vietas, the sum of the roots of both of these equations is 10 -10 .

But the question asks for sum of INTEGRAL values of x, and the solutions for x^2+5x+6 = 1 are not integers! So the correct answer should be -5!

Wei Chen - 4 years, 11 months ago

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Good point. I modified the phrasing of the problem. The use of "integral values of x" was added after the problem was published (check the reports of the problem).

Alex G - 4 years, 11 months ago

But 0 is neither positive nor negative so the answer should be -5.

Kevin Glentworth - 4 years, 10 months ago

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Good point. The wording has once again been modified.

Alex G - 4 years, 10 months ago
Joe Potillor
Dec 2, 2016

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