Question

In the following exercises, solve.

$2y^3+2y^2=12y$

Short answer

Answer is$y=0,y=2,y=-3$

Step-by-step solution

Step 1. Given

The equation is$2y^3+2y^2=12y$

Step 2. Simplify the equation

$$\begin{gathered} 2y^3+2y^2=12y \\ 2y^3+2y^2-12y=0 \\ {y}^3+y^2-6y=0 \\ y(y^2+y-6)=0 \\ y(y^2+3y-2y-6)=0 \\ y\left(y\right(y+3)-2(y+3\left)\right)=0 \\ y(y+3)(y-2)=0 \end{gathered}$$

So,

$y=0$

and

$$\begin{gathered} y+3=0 \\ y=-3 \end{gathered}$$

and

$$\begin{gathered} y-2=0 \\ y=2 \end{gathered}$$

Step 3. Check all the values

Put $y=0$

$$\begin{gathered} 2y^3+2y^2=12y \\ 2{\left(0\right)}^3+2{\left(0\right)}^2=12\left(0\right) \\ 0=0 \end{gathered}$$

True

Now put role="math" localid="1644656268781" $y=-3$in the original equation,

$$\begin{gathered} 2y^3+2y^2=12y \\ 2{(-3)}^3+2{(-3)}^2=12\times -3 \\ -54+18=-36 \\ -36=-36 \end{gathered}$$

True

Now put $y=2$in the original equation,

$$\begin{gathered} 2y^3+2y^2=12y \\ 2{\left(2\right)}^3+2{\left(2\right)}^2=12\times 2 \\ 16+8=24 \\ 24=24 \end{gathered}$$

True

Step 4. Required answer

Answer is$y=0,y=-3,y=2$