What are the solutions to the equation (x-2)^2(1-x)=-2 ?

The solutions to the equation (x-2)^2(1-x)=-2 are x=3,x=1+i,x=1-i

Find the zeros of (x-2)^2(1-x)=-2

The zeros of (x-2)^2(1-x)=-2 are x=3,x=1+i,x=1-i

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(x - 2)^{2} (1 - x) = - 2

x = 3, x = 1 + i, x = 1 - i

Solution steps

(x - 2)^{2} (1 - x) = - 2

Expand

(x - 2)^{2} (1 - x) : - x^{3} + 5 x^{2} - 8 x + 4

- x^{3} + 5 x^{2} - 8 x + 4 = - 2

Move

2

to the left side

- x^{3} + 5 x^{2} - 8 x + 6 = 0

Factor

- x^{3} + 5 x^{2} - 8 x + 6 : - (x - 3) (x^{2} - 2 x + 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 3 = 0 : x = 3

Solve

x^{2} - 2 x + 2 = 0 : x = 1 + i, x = 1 - i

The solutions are

x = 3, x = 1 + i, x = 1 - i

1000 (\frac{1}{2}) \cdot x^{28} = 20

s^{3} + 3 s^{2} + 25 + 3 = 0

15 x^{4} - 13 x^{2} - 2 = 0

x^{3} + x^{2} - 17 x + 15 = 0

2 a^{3} - 9 a^{2} - 4 = 0

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