Question 352757
3 consecutive numbers:  x, x+1, x+2
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sum is 1/5 their product:  x+(x+1)+(x+2)=1/5[x*(x+1)*(x+2)]
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 x+(x+1)+(x+2)=1/5[x*(x+1)*(x+2)]
{{{3x+3=(x^2+x)(x+2)/5}}}
{{{15x+15=(x^2+x)*(x+2)=x^3+2x^2+x^2+2x}}}
{{{0=x^3+3x^2-13x-15}}}

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since highest power is cubed, there are a total of 3 roots
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by descartes sign test,  there is one positive real root
possible real roots -+1,-+3,-+5,-+15
actual roots will rarely be in the extremes, so try near the middle, try 3
via synthetic division.
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{{{x^3+3x^2-13x-15}}} divided by root 3 yields {{{x^2+6x+5}}}
in other words
{{{x^3+3x^2-13x-15=(x-3)*(x^2+6x+5)}}}
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factor {{{x^2+6x+5}}}
(x+5)*(x+1)
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Therefore: {{{x^3+3x^2-13x-15=(x-3)*(x+5)*(x+1)}}} 
with potentially 3 solutions
if x=3, then x+1=4 and x+2=5
if x=-5 then x+1=-4 and x+2=-3
if x=-1 then x+1=0 and x+2=1

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(3,4,5) or (-5,-4,-3) or (-1,0,1)
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validate each set of solutions
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(3,4,5)
sum=3+4+5=12
1/5 product= 1/5*(3*4*5)=60/5=12
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(-5,-4,-3)
sum=-5-4-3=-12
1/5 product= 1/5*(-5)*(-4)*(-3)=-60/5=-12
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(1,0,-1)
sum=1+0-1=0
1/5 product = 1/5*(1)*(0)*(-1)=0/5=0
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so there exists more than one solution. All three of these solutions work.