What are the solutions to the equation 3x^3-4x^2+4x=16 ?

The solutions to the equation 3x^3-4x^2+4x=16 are x=2,x=-1/3+i(sqrt(23))/3 ,x=-1/3-i(sqrt(23))/3

Find the zeros of 3x^3-4x^2+4x=16

The zeros of 3x^3-4x^2+4x=16 are x=2,x=-1/3+i(sqrt(23))/3 ,x=-1/3-i(sqrt(23))/3

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

x = 2, x = - \frac{1}{3} + i \frac{23}{3}, x = - \frac{1}{3} - i \frac{23}{3}

Solution steps

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

Move

16

to the left side

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

Factor

3 x^{3} - 4 x^{2} + 4 x - 16 : (x - 2) (3 x^{2} + 2 x + 8)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

3 x^{2} + 2 x + 8 = 0 : x = - \frac{1}{3} + i \frac{23}{3}, x = - \frac{1}{3} - i \frac{23}{3}

The solutions are

x = 2, x = - \frac{1}{3} + i \frac{23}{3}, x = - \frac{1}{3} - i \frac{23}{3}

(3 - x)^{5} + (x - 5)^{5} = - 32

c, c^{1} c^{2} = r^{1} + r^{2}

x^{7} = 120

x^{4} - 2 = 0

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

What are the solutions to the equation 3x^3-4x^2+4x=16 ?
The solutions to the equation 3x^3-4x^2+4x=16 are x=2,x=-1/3+i(sqrt(23))/3 ,x=-1/3-i(sqrt(23))/3
Find the zeros of 3x^3-4x^2+4x=16
The zeros of 3x^3-4x^2+4x=16 are x=2,x=-1/3+i(sqrt(23))/3 ,x=-1/3-i(sqrt(23))/3