Q. 312 - Intermediate Algebra

Question

Solve quadratic equations by factoring.

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

Step-by-Step Solution

Answer

The roots are $0, 2, -3$.

Step 1. Rearrange the terms

Rearranging the terms to bring them on a single side,

$2y^3+2y^2-12y=0$

Step 2. Factor the greatest common factor

Factoring out the greatest common factor first,

$2y(y^2+y-6)=0$

Step 3. Simplify the quadratic equation

We factor the trinomial first,

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

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

when $2y=0$,

$y=0$.

When $y+3=0$,

$y=-3$.

When $y-2=0$,

$y=2$.

Step 4. Check

Resubstitute each of the roots separately into the original equation.

When $y=0$,

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

This is true.

When $y=-3$,

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

This is true.

When $y=2$,

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

This is true as well.

Thus, all the roots satisfy the original equation.