What are the solutions to the equation 0=2x^3-x^2-8x+4 ?

The solutions to the equation 0=2x^3-x^2-8x+4 are x= 1/2 ,x=-2,x=2

Find the zeros of 0=2x^3-x^2-8x+4

The zeros of 0=2x^3-x^2-8x+4 are x= 1/2 ,x=-2,x=2

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0 = 2 x^{3} - x^{2} - 8 x + 4

x = \frac{1}{2}, x = - 2, x = 2

Decimal Notation

x = 0.5, x = - 2, x = 2

Solution steps

0 = 2 x^{3} - x^{2} - 8 x + 4

Switch sides

2 x^{3} - x^{2} - 8 x + 4 = 0

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 x - 1 = 0 : x = \frac{1}{2}

Solve

x + 2 = 0 : x = - 2

Solve

x - 2 = 0 : x = 2

The solutions are

x = \frac{1}{2}, x = - 2, x = 2

2 x^{3} + 2 x = 0

- x^{3} + 7 x^{2} - 2 x - 40 = 0

x^{4} = 5 x^{2} - 4

4 x^{4} + 4 = 5 x^{2}

(x + 2)^{3} = 27

What are the solutions to the equation 0=2x^3-x^2-8x+4 ?
The solutions to the equation 0=2x^3-x^2-8x+4 are x= 1/2 ,x=-2,x=2
Find the zeros of 0=2x^3-x^2-8x+4
The zeros of 0=2x^3-x^2-8x+4 are x= 1/2 ,x=-2,x=2