Q. 311 - Intermediate Algebra

Question

Solve quadratic equations by factoring.

$36x^3+24x^2=-4x$

Step-by-Step Solution

Answer

The roots are $0, -\frac{1}{3}$.

Step 1. Rearrange the terms

Rearranging the terms to bring them on a single side,

$36x^3+24x^2+4x=0$

Step 2. Factor the greatest common factor

Factoring out the greatest common factor first,

$4x(9x^2+6x+1)=0$

Step 3. Simplify the quadratic equation

We factor the trinomial first,

$$\begin{gathered} 4x(9x^2+6x+1)=0 \\ 4x(9x^2+3x+3x+1)=0 \\ 4x\left(3x\right(3x+1)+1(3x+1\left)\right)=0 \\ 4x(3x+1)(3x+1)=0 \\ 4x{(3x+1)}^2=0 \end{gathered}$$

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

when $4x=0$,

$x=0$

when $3x+1=0$,

$x=-\frac{1}{3}$

Step 4. Check

Resubstitute each of the roots separately into the original equation.

When $x=0$,

$$\begin{gathered} 36x^3+24x^2=-4x \\ 36{\left(0\right)}^3+24{\left(0\right)}^2=-4\left(0\right) \\ 0+0=0 \\ 0=0 \end{gathered}$$

This is true.

When $x=-\frac{1}{3}$,

$$\begin{gathered} 36x^3+24x^2=-4x \\ 36\left({(-\frac{1}{3})}^3\right)+24\left({(-\frac{1}{3})}^2\right)=-4(-\frac{1}{3}) \\ 36(-\frac{1}{27})+24\left(\frac{1}{9}\right)=\frac{4}{3} \\ -\frac{4}{3}+\frac{8}{3}=\frac{4}{3} \\ \frac{4}{3}=\frac{4}{3} \end{gathered}$$

This is also true.

Thus, both roots satisfy the original equation.