What are the solutions to the equation 12y^3+24y^2=3y+6 ?

The solutions to the equation 12y^3+24y^2=3y+6 are y=-2,y=-1/2 ,y= 1/2

Find the zeros of 12y^3+24y^2=3y+6

The zeros of 12y^3+24y^2=3y+6 are y=-2,y=-1/2 ,y= 1/2

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12 y^{3} + 24 y^{2} = 3 y + 6

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

Decimal Notation

y = - 2, y = - 0.5, y = 0.5

Solution steps

12 y^{3} + 24 y^{2} = 3 y + 6

Move

6

to the left side

12 y^{3} + 24 y^{2} - 6 = 3 y

Move

3 y

to the left side

12 y^{3} + 24 y^{2} - 6 - 3 y = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

12 y^{3} + 24 y^{2} - 3 y - 6 = 0

Factor

12 y^{3} + 24 y^{2} - 3 y - 6 : 3 (y + 2) (2 y + 1) (2 y - 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

y + 2 = 0 : y = - 2

Solve

2 y + 1 = 0 : y = - \frac{1}{2}

Solve

2 y - 1 = 0 : y = \frac{1}{2}

The solutions are

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

x (x + 10) (x^{2} - 5) = 0

24 x^{4} - 8 x^{2} + 30 = 0

x^{3} + 3 x^{2} - 14 x - 42 = 0

4 x^{4} = 13 x^{2} - 9

a^{9} = 17 + (9 - 1) - 8

What are the solutions to the equation 12y^3+24y^2=3y+6 ?
The solutions to the equation 12y^3+24y^2=3y+6 are y=-2,y=-1/2 ,y= 1/2
Find the zeros of 12y^3+24y^2=3y+6
The zeros of 12y^3+24y^2=3y+6 are y=-2,y=-1/2 ,y= 1/2