What are the solutions to the equation 2x^3-4x^2-2x+4=0 ?

The solutions to the equation 2x^3-4x^2-2x+4=0 are x=2,x=-1,x=1

Find the zeros of 2x^3-4x^2-2x+4=0

The zeros of 2x^3-4x^2-2x+4=0 are x=2,x=-1,x=1

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

x = 2, x = - 1, x = 1

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 2 = 0 : x = 2

Solve

x + 1 = 0 : x = - 1

Solve

x - 1 = 0 : x = 1

The solutions are

x = 2, x = - 1, x = 1

n - 70 = 200 - (15 x^{4})

x, x^{3} - y^{3} = 8

x, x^{3} - y^{3} = 1

3 x^{3} + 25 = 1

7 = 56 x^{3}

What are the solutions to the equation 2x^3-4x^2-2x+4=0 ?
The solutions to the equation 2x^3-4x^2-2x+4=0 are x=2,x=-1,x=1
Find the zeros of 2x^3-4x^2-2x+4=0
The zeros of 2x^3-4x^2-2x+4=0 are x=2,x=-1,x=1