Question

Find the values of y:

3y^3(y^2+ 2/3y)=0

Answer 1

y = 0 or y = -2/3

Explanation:

`3y^3\left(y^2+(2)/(3)y\right)=0`

The product is zero if one of the factors is equal to zero. Therefore

`3y^3\left(y^2+(2)/(3)y\right)=0\iff3y^3=0\ \vee\ y^2+(2)/(3)y=0\\\\3y^3=0\qquad\text{divide both sides by 3}\\\\y^3=0\to \boxed{y=0}\\\\y^2+(2)/(3)y=0\\\\y\left(y+(2)/(3)\right)=0\iff y=0\ \vee\ y+(2)/(3)=0\to \boxed{y=-(2)/(3)}`

Answer 2

y =0 y = -2/3

Explanation:

3y^3(y^2+ 2/3y)=0

Factor out a y

3y^3 *y (y+2/3) =0

3y^4 (y+2/3) =0

Using the zero product property

3y^4 =0 y+2/3 =0

y^4 = 0 y+2/3-2/3 =0

y =0 y = -2/3