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

The solutions to the equation (2x-2)^2(x-2)=(2x-2)(x-2)^2 are x=0,x=1,x=2

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

The zeros of (2x-2)^2(x-2)=(2x-2)(x-2)^2 are x=0,x=1,x=2

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

x = 0, x = 1, x = 2

Solution steps

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

Move

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

to the left side

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

Factor

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

2 x (x - 1) (x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 2 = 0 : x = 2

The solutions are

x = 0, x = 1, x = 2

x^{5} + 8 x^{3} = 0

0.001 x^{3} - 2 x + 30 = 0

4 x^{9} = 12 x^{3}

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

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

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