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

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

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

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

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

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

Solution steps

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

Factor

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

(2 x - 1) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x - 1 = 0 or x^{2} + 4 = 0

Solve

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

Solve

x^{2} + 4 = 0 : x = 2 i, x = - 2 i

The solutions are

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

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

5 s^{3} + 7 s + 6.4 = 0

a^{4} = 10 a^{2}

2 (x^{2} - 2 x)^{2} - 5 (x^{2} - 2 x) = 7

z^{6} + z + 1 = 0

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