(x^3-4x^2+5x-20)/(x-4)

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Solution for (x^3-4x^2+5x-20)/(x-4) equation: 


D( x )

x-4 = 0

x-4 = 0

x-4 = 0

x-4 = 0 // + 4

x = 4

x in (-oo:4) U (4:+oo)

(x^3-(4*x^2)+5*x-20)/(x-4) = 0

(x^3-4*x^2+5*x-20)/(x-4) = 0

x^3-4*x^2+5*x-20 = 0

x^3-4*x^2+5*x-20

x^2*(x-4)+5*x-20

5*(x-4)+x^2*(x-4)

(x^2+5)*(x-4)

x^2+5 = 0

1*x^2 = -5 // : 1

x^2 = -5

x belongs to the empty set

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