What are the solutions to the equation (x)=3x^3-12x^2+8x+2 ?

The solutions to the equation (x)=3x^3-12x^2+8x+2 are x=1,x=(9+sqrt(105))/6 ,x=(9-sqrt(105))/6

Find the zeros of (x)=3x^3-12x^2+8x+2

The zeros of (x)=3x^3-12x^2+8x+2 are x=1,x=(9+sqrt(105))/6 ,x=(9-sqrt(105))/6

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(x) = 3 x^{3} - 12 x^{2} + 8 x + 2

x = 1, x = \frac{9 + 105}{6}, x = \frac{9 - 105}{6}

Solution steps

(x) = 3 x^{3} - 12 x^{2} + 8 x + 2

Refine

x = 3 x^{3} - 12 x^{2} + 8 x + 2

Switch sides

3 x^{3} - 12 x^{2} + 8 x + 2 = x

Move

x

to the left side

3 x^{3} - 12 x^{2} + 7 x + 2 = 0

Factor

3 x^{3} - 12 x^{2} + 7 x + 2 : (x - 1) (3 x^{2} - 9 x - 2)

(x - 1) (3 x^{2} - 9 x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 3 x^{2} - 9 x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

3 x^{2} - 9 x - 2 = 0 : x = \frac{9 + 105}{6}, x = \frac{9 - 105}{6}

The solutions are

x = 1, x = \frac{9 + 105}{6}, x = \frac{9 - 105}{6}

- x^{2} (x - 2) (2 - x) = 0

x^{3} - 5 = - 7

x^{4} - 8 x^{2} + 9 = 0

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

500 \cdot b^{3} = 10

What are the solutions to the equation (x)=3x^3-12x^2+8x+2 ?
The solutions to the equation (x)=3x^3-12x^2+8x+2 are x=1,x=(9+sqrt(105))/6 ,x=(9-sqrt(105))/6
Find the zeros of (x)=3x^3-12x^2+8x+2
The zeros of (x)=3x^3-12x^2+8x+2 are x=1,x=(9+sqrt(105))/6 ,x=(9-sqrt(105))/6