What are the solutions to the equation y^3+3y^2=4y+12 ?

The solutions to the equation y^3+3y^2=4y+12 are y=-3,y=-2,y=2

Find the zeros of y^3+3y^2=4y+12

The zeros of y^3+3y^2=4y+12 are y=-3,y=-2,y=2

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y^{3} + 3 y^{2} = 4 y + 12

y = - 3, y = - 2, y = 2

Solution steps

y^{3} + 3 y^{2} = 4 y + 12

Move

12

to the left side

y^{3} + 3 y^{2} - 12 = 4 y

Move

4 y

to the left side

y^{3} + 3 y^{2} - 12 - 4 y = 0

Write in the standard form

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

y^{3} + 3 y^{2} - 4 y - 12 = 0

Factor

y^{3} + 3 y^{2} - 4 y - 12 : (y + 3) (y + 2) (y - 2)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

y + 3 = 0 : y = - 3

Solve

y + 2 = 0 : y = - 2

Solve

y - 2 = 0 : y = 2

The solutions are

y = - 3, y = - 2, y = 2

x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6 = 0

8 x^{3} - 1 = 0

w^{3} = i

z^{3} = \frac{i}{8}

x^{2} (x - 6) = 0

What are the solutions to the equation y^3+3y^2=4y+12 ?
The solutions to the equation y^3+3y^2=4y+12 are y=-3,y=-2,y=2
Find the zeros of y^3+3y^2=4y+12
The zeros of y^3+3y^2=4y+12 are y=-3,y=-2,y=2