Q. 309 - Intermediate Algebra

Question

Solve quadratic equations by factoring.

$2x^3+72x=24x^2$

Step-by-Step Solution

Answer

The roots are $0, 6$.

Step 1. Rearrange the terms

Rearranging the terms to bring them on a single side,

$2x^3-24x^2+72x=0$

Step 2. Factor the greatest common factor

Factoring out the greatest common factor first,

$2x(x^2-12x+36)=0$

Step 3. Simplify the quadratic equation

We factor the trinomial first,

$$\begin{gathered} 2x(x^2-12x+36)=0 \\ 2x(x^2-6x-6x+36)=0 \\ 2x\left(x\right(x-6)-6(x-6\left)\right)=0 \\ 2x(x-6)(x-6)=0 \end{gathered}$$

Using the Zero Product Property to set each factor to $0$,

when $2x=0$,

$x=0$

when $x-6=0$,

$x=6$

Step 4. Check

Resubstitute each of the roots separately into the original equation.

When $x=0$,

$$\begin{gathered} 2x^3+72x=24x^2 \\ 2{\left(0\right)}^3+72\left(0\right)=24\left(0\right) \\ 0=0 \end{gathered}$$

When $x=6$,

$$\begin{gathered} 2x^3+72x=24x^2 \\ 2{\left(6\right)}^3+72\left(6\right)=24{\left(6\right)}^{2}432+432=864 \\ 864=864 \end{gathered}$$

Thus, both roots satisfy the original equation.