What are the solutions to the equation x^3-3x^2+6x-4=0 ?

The solutions to the equation x^3-3x^2+6x-4=0 are x=1,x=1+sqrt(3)i,x=1-sqrt(3)i

Find the zeros of x^3-3x^2+6x-4=0

The zeros of x^3-3x^2+6x-4=0 are x=1,x=1+sqrt(3)i,x=1-sqrt(3)i

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

x = 1, x = 1 + 3 i, x = 1 - 3 i

Solution steps

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

Factor

x^{3} - 3 x^{2} + 6 x - 4 : (x - 1) (x^{2} - 2 x + 4)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x - 1 = 0 : x = 1

Solve

x^{2} - 2 x + 4 = 0 : x = 1 + 3 i, x = 1 - 3 i

The solutions are

x = 1, x = 1 + 3 i, x = 1 - 3 i

4 (x^{2} + 9)^{2} = 0.24 x

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

x^{4} + 7 x^{2} - 18 = 0

7 x = x^{6} + 5 (3 x + 3) - x

x^{3} - 2 x = 1

What are the solutions to the equation x^3-3x^2+6x-4=0 ?
The solutions to the equation x^3-3x^2+6x-4=0 are x=1,x=1+sqrt(3)i,x=1-sqrt(3)i
Find the zeros of x^3-3x^2+6x-4=0
The zeros of x^3-3x^2+6x-4=0 are x=1,x=1+sqrt(3)i,x=1-sqrt(3)i