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

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

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

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

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2 - x = 2 (x - 2)^{3}

x = 2, x = 2 + i \frac{2}{2}, x = 2 - i \frac{2}{2}

Solution steps

2 - x = 2 (x - 2)^{3}

Expand

2 (x - 2)^{3} : 2 x^{3} - 12 x^{2} + 24 x - 16

2 - x = 2 x^{3} - 12 x^{2} + 24 x - 16

Switch sides

2 x^{3} - 12 x^{2} + 24 x - 16 = 2 - x

Move

x

to the left side

2 x^{3} - 12 x^{2} + 25 x - 16 = 2

Move

2

to the left side

2 x^{3} - 12 x^{2} + 25 x - 18 = 0

Factor

2 x^{3} - 12 x^{2} + 25 x - 18 : (x - 2) (2 x^{2} - 8 x + 9)

(x - 2) (2 x^{2} - 8 x + 9) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

2 x^{2} - 8 x + 9 = 0 : x = 2 + i \frac{2}{2}, x = 2 - i \frac{2}{2}

The solutions are

x = 2, x = 2 + i \frac{2}{2}, x = 2 - i \frac{2}{2}

x^{4} = - 16

(- 1) (6 x^{3} + x^{2} + x - 1) = 0

y, 27 x^{3} + y^{3} + z^{3} = 9 x \cdot z

13 3^{5} + 11 0^{5} + 8 4^{5} + 2 7^{5} = n^{5}

r^{3} = 8

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