What are the solutions to the equation y^4-y^3-3y^2+5y^1-2y=0 ?

The solutions to the equation y^4-y^3-3y^2+5y^1-2y=0 are y=0,y=1,y=-sqrt(3),y=sqrt(3)

Find the zeros of y^4-y^3-3y^2+5y^1-2y=0

The zeros of y^4-y^3-3y^2+5y^1-2y=0 are y=0,y=1,y=-sqrt(3),y=sqrt(3)

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Popular >

y^{4} - y^{3} - 3 y^{2} + 5 y^{1} - 2 y = 0

y = 0, y = 1, y = - 3, y = 3

Decimal Notation

y = 0, y = 1, y = - 1.73205 \dots, y = 1.73205 \dots

Solution steps

y^{4} - y^{3} - 3 y^{2} + 5 y^{1} - 2 y = 0

Factor

y^{4} - y^{3} - 3 y^{2} + 5 y^{1} - 2 y : y (y - 1) (y + 3) (y - 3)

y (y - 1) (y + 3) (y - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or y - 1 = 0 or y + 3 = 0 or y - 3 = 0

Solve

y - 1 = 0 : y = 1

Solve

y + 3 = 0 : y = - 3

Solve

y - 3 = 0 : y = 3

The solutions are

y = 0, y = 1, y = - 3, y = 3

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

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

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

2 w^{4} - 5 w^{2} + 2 = 0

m^{3} - 7 m + 6 = 0

What are the solutions to the equation y^4-y^3-3y^2+5y^1-2y=0 ?
The solutions to the equation y^4-y^3-3y^2+5y^1-2y=0 are y=0,y=1,y=-sqrt(3),y=sqrt(3)
Find the zeros of y^4-y^3-3y^2+5y^1-2y=0
The zeros of y^4-y^3-3y^2+5y^1-2y=0 are y=0,y=1,y=-sqrt(3),y=sqrt(3)