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

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

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

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

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

x = 4, x = \frac{1}{2} i, x = - \frac{1}{2} i

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 4 = 0 : x = 4

Solve

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

The solutions are

x = 4, x = \frac{1}{2} i, x = - \frac{1}{2} i

4 x^{3} - 12 x^{2}

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

3 x^{3} + 5 x^{2} + 75 x + 125 = 0

m^{3} - 3 m + 2 = 0

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 = 0

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