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

The solutions to the equation 3y^3-5y^2-2y=0 are y=0,y=-1/3 ,y=2

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

The zeros of 3y^3-5y^2-2y=0 are y=0,y=-1/3 ,y=2

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3 y^{3} - 5 y^{2} - 2 y = 0

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

Decimal Notation

y = 0, y = - 0.33333 \dots, y = 2

Solution steps

3 y^{3} - 5 y^{2} - 2 y = 0

Factor

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

y (3 y + 1) (y - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or 3 y + 1 = 0 or y - 2 = 0

Solve

3 y + 1 = 0 : y = - \frac{1}{3}

Solve

y - 2 = 0 : y = 2

The solutions are

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

4 x^{6} - 64 x^{2} = 0

x^{4} - 2 x^{3} + 8 x^{2} - 32 x - 128 = 0

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

x^{3} = 2^{30}

x^{3} - 16 x^{2} + 55 x + 72 = 0

What are the solutions to the equation 3y^3-5y^2-2y=0 ?
The solutions to the equation 3y^3-5y^2-2y=0 are y=0,y=-1/3 ,y=2
Find the zeros of 3y^3-5y^2-2y=0
The zeros of 3y^3-5y^2-2y=0 are y=0,y=-1/3 ,y=2