What are the solutions to the equation 0=y^4-y ?

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

Find the zeros of 0=y^4-y

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

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0 = y^{4} - y

y = 0, y = 1, y = - \frac{1}{2} + \frac{3}{2} i, y = - \frac{1}{2} - \frac{3}{2} i

Solution steps

0 = y^{4} - y

Switch sides

y^{4} - y = 0

Factor

y^{4} - y : y (y^{3} - 1)

y (y^{3} - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or y^{3} - 1 = 0

Solve

y^{3} - 1 = 0 : y = 1, y = - \frac{1}{2} + \frac{3}{2} i, y = - \frac{1}{2} - \frac{3}{2} i

The solutions are

y = 0, y = 1, y = - \frac{1}{2} + \frac{3}{2} i, y = - \frac{1}{2} - \frac{3}{2} i

- 7 x - 8 + x^{4} = 0

9 x^{3} + 9 = 0

x^{3} + 5 x^{2} + 2 x - 8 = x^{2} + 3 x - 4

(3 w - 21) (w^{2} + 9 w + 20) = 0

y, 34 x - y^{3} = 23 x - 9

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