(x^3-3x^2+5x-3)/(x-1)

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Solution for (x^3-3x^2+5x-3)/(x-1) equation: 


D( x )

x-1 = 0

x-1 = 0

x-1 = 0

x-1 = 0 // + 1

x = 1

x in (-oo:1) U (1:+oo)

(x^3-(3*x^2)+5*x-3)/(x-1) = 0

(x^3-3*x^2+5*x-3)/(x-1) = 0

x^3-3*x^2+5*x-3 = 0

x^3-3*x^2+5*x-3 = 0

{ 1, -1, 3, -3 }

1

x = 1

x^3-3*x^2+5*x-3 = 0

1

x-1

x^2-2*x+3

x^3-3*x^2+5*x-3

x-1

x^2-x^3

5*x-2*x^2-3

2*x^2-2*x

3*x-3

3-3*x

0

x^2-2*x+3 = 0

DELTA = (-2)^2-(1*3*4)

DELTA = -8

DELTA < 0

x in { 1} 

x-1 = 0

1 = 0

x belongs to the empty set

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