(w^3-4w^2+7w-12)/(w-3)

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Solution for (w^3-4w^2+7w-12)/(w-3) equation:


D( w )

w-3 = 0

w-3 = 0

w-3 = 0

w-3 = 0 // + 3

w = 3

w in (-oo:3) U (3:+oo)

(w^3-(4*w^2)+7*w-12)/(w-3) = 0

(w^3-4*w^2+7*w-12)/(w-3) = 0

w^3-4*w^2+7*w-12 = 0

w^3-4*w^2+7*w-12 = 0

{ 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12 }

1

w = 1

w^3-4*w^2+7*w-12 = -8

1

-1

w = -1

w^3-4*w^2+7*w-12 = -24

-1

2

w = 2

w^3-4*w^2+7*w-12 = -6

2

-2

w = -2

w^3-4*w^2+7*w-12 = -50

-2

3

w = 3

w^3-4*w^2+7*w-12 = 0

3

w-3

w^2-w+4

w^3-4*w^2+7*w-12

w-3

3*w^2-w^3

7*w-w^2-12

w^2-3*w

4*w-12

12-4*w

0

w^2-w+4 = 0

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

DELTA = -15

DELTA < 0

w in { 3}

w-3 = 0

1 = 0

w belongs to the empty set

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