Chapter 4.2, Problem 72AYU

To determine

To find: The real solutions of $f$.

Answer to Problem 72AYU

$\text{The real solution }=\frac{2}{3}$.

Explanation of Solution

Given:

$f\left(x\right)=x^3-\frac{2}{3}{x}^2+3x-2=0$

$F$ is a polynomial function.

The degree of the polynomial function is 3. Therefore the number of real zeros by real zero theorem can be at most $=3$.

From Descartes’ rule of signs

$f\left(x\right)=x^3-\frac{2}{3}{x}^2+3x-2$

$+$ To $-$ $-$ to $+$ $+$ to $-$

There will be 3 or 1 positive real zeros.

Rational zeros theorem provides information about the potential rational zeros of a polynomial function with integer coefficients.

If $\frac{p}{q}$ in its lowest terms is a rational zero of $f$, then $p$ is a factor of $a_0$ and $q$ is the factor of $a_{\mathrm{n}}$.

$f\left(x\right)=x^3-\frac{2}{3}{x}^2+3x-2=0,f\left(x\right)=3x^3-2x^2+9x-6=0$

Here $a_0=-6$ and $a_{\mathrm{n}}=3$

Zeros of 3, $q=\pm 1,\pm 3$

Zeros of $-6$, $p=\pm 1,\pm 2,\pm 3,\pm 6$

The potential rational zeros of $f=\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3}$.

$x=\frac{2}{3}$ is zero of $f$.

Use long division or Synthetic division of $3x^3-2x^2+9x-6=0$.

$f\left(x\right)=\left(x-\frac{2}{3}\right)\left(3x^2+9\right)$

The depressed equation $=3\left(x^2+3\right), x=\pm i\sqrt{3}$

The real zeros $=\frac{2}{3}$.