Question

Solve quadratic equations by factoring.

$3y^3+48y=24y^2$

Short answer

The roots are$0,4$.

Step-by-step solution

Step 1. Rearrange the terms 

Rearranging the terms to bring them on a single side,

$3y^3-24y^2+48y=0$

Step 2. Factor the greatest common factor

Factoring out the greatest common factor first,

$3y(y^2-8y+16)=0$

Step 3. Simplify the quadratic equation

We factor the trinomial first,

$$\begin{gathered} 3y(y^2-8y+16)=0 \\ 3y(y^2-4y-4y+16)=0 \\ 3y\left(y\right(y-4)-4(y-4\left)\right)=0 \\ 3y(y-4)(y-4)=0 \end{gathered}$$

Use the Zero Product Property to set each factor to $0$, we get,

when $3y=0$,

$y=0$

when $y-4=0$

$y=4$

Step 4. Check

Resubstitute each of the roots separately into the original equation.

When $y=0$,

$$\begin{gathered} 3y^3+48y=24y^2 \\ 3{\left(0\right)}^3+48\left(0\right)=24\left(0\right) \\ 0=0 \end{gathered}$$

This is true.

When $y=4$,

localid="1661950054589" $$\begin{gathered} 3y^3+48y=24y^2 \\ 3{\left(4\right)}^3+48\left(4\right)=24{\left(4\right)}^2 \\ 192+192=384 \\ 384=384 \end{gathered}$$

This is also true.

Thus, both roots satisfy the original equation.