solve for factorial.(n+2)!/(n-1)!=336

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beckden | High School Teacher | (Level 1) Educator

Posted on

The solutions to n^3+3n^2+2n-336=0 has 6 as a root. so

n=6 is a solution


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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to solve for n the factorial equation:


You should know that `(n+2)! = (n-1)!*n*(n+1)*(n+2)`


Reducing by (n-1)! yields:

`n*(n+1)*(n+2) = 336`

Opening the brackets yields:

`n^3 + 3n^2 + 2n - 336 = 0`

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