(n^3+n-10)/(n-2)

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Solution for (n^3+n-10)/(n-2) equation: 


D( n )

n-2 = 0

n-2 = 0

n-2 = 0

n-2 = 0 // + 2

n = 2

n in (-oo:2) U (2:+oo)

(n^3+n-10)/(n-2) = 0

n^3+n-10 = 0

n^3+n-10 = 0

{ 1, -1, 2, -2, 5, -5, 10, -10 }

1

n = 1

n^3+n-10 = -8

1

-1

n = -1

n^3+n-10 = -12

-1

2

n = 2

n^3+n-10 = 0

2

n-2

n^2+2*n+5

n^3+n-10

n-2

2*n^2-n^3

2*n^2+n-10

4*n-2*n^2

5*n-10

10-5*n

0

n^2+2*n+5 = 0

DELTA = 2^2-(1*4*5)

DELTA = -16

DELTA < 0

n in { 2} 

n-2 = 0

1 = 0

n belongs to the empty set

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