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

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

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

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

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

x = 0, x = \frac{3}{2} + i \frac{3}{2}, x = \frac{3}{2} - i \frac{3}{2}

Solution steps

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

Switch sides

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x^{2} - 3 x + 3 = 0 : x = \frac{3}{2} + i \frac{3}{2}, x = \frac{3}{2} - i \frac{3}{2}

The solutions are

x = 0, x = \frac{3}{2} + i \frac{3}{2}, x = \frac{3}{2} - i \frac{3}{2}

x^{10} + x^{5} + 1 = 0

y^{3} - 25 y^{2} - 178 y + 1165 = 0

x^{4} - 52 x^{2} + 147 = 0

x^{4} - 3 = 2 x + 1

x, g = r (x + q)^{7} + s

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