Question

Find the real solutions of each equation.

$x^3-\frac{2}{3}{x}^2+3x-2=0$

Short answer

The only real solution of the equation is $x=\frac{2}{3}$.

Step-by-step solution

Step 1. Write as integer coefficient 

Multiply both sides of the equation by $3$.

role="math" localid="1646222419003" $\begin{array}{rcl}3(x^3-\frac{2}{3}{x}^2+3x-2)& =& 3·0\\ 3x^3-2x^2+9x-6& =& 0\end{array}$

Now the corresponding function is $f\left(x\right)=\begin{array}{rcl}& & 3\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & 3\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 9\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 6\end{array}$.

Step 2. Use the rational zero theorem   

Now all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $6$are

$p:\pm 1,\pm 2,\pm 3,\pm 6$

The factors of the leading coefficient $3$are

$q:\pm 1,\pm 3$

So the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3}$

Step 3. Find function values 

For the function $f\left(x\right)=\begin{array}{rcl}& & 3\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & {}^3\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 9\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 6\end{array}$, the function value $f\left(\frac{1}{3}\right)$is given as

role="math" localid="1646222478706" $$\begin{gathered} f\left(\frac{1}{3}\right)=\begin{array}{rcl}& & 3\end{array}{\left(\frac{1}{3}\right)}^3-2{\left(\frac{1}{3}\right)}^2+9\left(\frac{1}{3}\right)\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 6\end{array} \\ f\left(\frac{1}{3}\right)=\frac{1}{9}-\frac{2}{9}+3-6 \\ f\left(\frac{1}{3}\right)=-\frac{1}{9}-3 \\ f\left(\frac{1}{3}\right)=-\frac{28}{9} \end{gathered}$$

The function value $f\left(\frac{2}{3}\right)$is given as

role="math" localid="1646222496784" $$\begin{gathered} f\left(\frac{2}{3}\right)=\begin{array}{rcl}& & 3\end{array}{\left(\frac{2}{3}\right)}^3\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 2\end{array}{\left(\frac{2}{3}\right)}^2\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 9\end{array}\left(\frac{2}{3}\right)\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 6\end{array} \\ f\left(\frac{2}{3}\right)=\frac{8}{9}-\frac{8}{9}+6-6 \\ f\left(\frac{2}{3}\right)=0 \end{gathered}$$

Step 4. Perform synthetic division 

As $f\left(\frac{2}{3}\right)=0$, so $x=\frac{2}{3}$is a zero.

So on performing synthetic division we get

So the depressed polynomial is $x^2+3$. The polynomial is prime and it does not have any real solution.

So the only real solution is $x=\frac{2}{3}$.