What are the solutions to the equation 12(x^3-4x^2+x+6)=0 ?

The solutions to the equation 12(x^3-4x^2+x+6)=0 are x=-1,x=2,x=3

Find the zeros of 12(x^3-4x^2+x+6)=0

The zeros of 12(x^3-4x^2+x+6)=0 are x=-1,x=2,x=3

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12 (x^{3} - 4 x^{2} + x + 6) = 0

x = - 1, x = 2, x = 3

Solution steps

12 (x^{3} - 4 x^{2} + x + 6) = 0

Factor

12 (x^{3} - 4 x^{2} + x + 6) : 12 (x + 1) (x - 2) (x - 3)

12 (x + 1) (x - 2) (x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 1 = 0 or x - 2 = 0 or x - 3 = 0

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

x - 3 = 0 : x = 3

The solutions are

x = - 1, x = 2, x = 3

3 x^{3} - 8 x^{2} + x + 2 = 0

2 x^{3} - 11 x = - 9

(4 x^{3} + 12 x^{2} - 1) = 0

a^{6} = 16 a^{20}, a^{4} = 10

- 4 x^{3} + 16 x^{2} - 9 x - 9 = 0

What are the solutions to the equation 12(x^3-4x^2+x+6)=0 ?
The solutions to the equation 12(x^3-4x^2+x+6)=0 are x=-1,x=2,x=3
Find the zeros of 12(x^3-4x^2+x+6)=0
The zeros of 12(x^3-4x^2+x+6)=0 are x=-1,x=2,x=3