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

The solutions to the equation x^4+2x^3-6x^2+3x=0 are x=0,x=1,x=(-3+sqrt(21))/2 ,x=(-3-sqrt(21))/2

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

The zeros of x^4+2x^3-6x^2+3x=0 are x=0,x=1,x=(-3+sqrt(21))/2 ,x=(-3-sqrt(21))/2

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

x = 0, x = 1, x = \frac{- 3 + 21}{2}, x = \frac{- 3 - 21}{2}

Decimal Notation

x = 0, x = 1, x = 0.79128 \dots, x = - 3.79128 \dots

Solution steps

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

Factor

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

x (x - 1) (x^{2} + 3 x - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or x^{2} + 3 x - 3 = 0

Solve

x - 1 = 0 : x = 1

Solve

x^{2} + 3 x - 3 = 0 : x = \frac{- 3 + 21}{2}, x = \frac{- 3 - 21}{2}

The solutions are

x = 0, x = 1, x = \frac{- 3 + 21}{2}, x = \frac{- 3 - 21}{2}

x^{4} + 9 x^{3} + 18 x^{2} = 0

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

x^{3} + 8 = - 56

12 x^{4} + 11 x^{3} - 26 x^{2} + x + 2 = 0

x, y = (x - 1)^{3}

What are the solutions to the equation x^4+2x^3-6x^2+3x=0 ?
The solutions to the equation x^4+2x^3-6x^2+3x=0 are x=0,x=1,x=(-3+sqrt(21))/2 ,x=(-3-sqrt(21))/2
Find the zeros of x^4+2x^3-6x^2+3x=0
The zeros of x^4+2x^3-6x^2+3x=0 are x=0,x=1,x=(-3+sqrt(21))/2 ,x=(-3-sqrt(21))/2