Chapter 4.2, Problem 66AYU

To determine

To find: The real solutions of $f$.

Answer to Problem 66AYU

$\text{The real solution = }\frac{5}{2}$.

Explanation of Solution

Given:

$f\left(x\right)=2x^3-3x^2-3x-5=0$

$F$ 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

$f\left(x\right)=2x^3-3x^2-3x-5$

$+$ To $-$

There will be 1 positive real zero.

$f\left(-x\right)=-2x^3-3x^2+3x-5$

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

There will be 2 or 0 negative 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 $a_{\mathrm{n}}$.

$f\left(x\right)=2x^3-3x^2-3x-5$

Here $a_0=-5$ and $a_{\mathrm{n}}=2$

Zeros of 2, $q=\pm 1,\pm 2$

Zeros of $-5$, $p=\pm 1,\pm 5$

The potential rational zeros of $f= \pm 1,\pm 2,\pm 5,\pm \frac{1}{2},\pm \frac{5}{2}$.

$x=\frac{5}{2}$ is zero of $f$.

Use long division or Synthetic division of $2x^3-3x^2-3x-5$.

$f\left(x\right)=\left(x-\frac{5}{2}\right)\left(2x^2+2x+2\right)$

The depressed equation $=2\left(x^2+x+1\right)$

$\left(x^2+x+1\right)$

Factorized further

$a=1$, $b=1$ and $c=1$

$x=\frac{-b\pm \sqrt[]{b^2-4\cdot a\cdot c}}{2\cdot a}=\frac{-1\pm \sqrt[]{1^2-4\cdot 1\cdot \left(1\right)}}{2\cdot 1}=\frac{-1\pm \sqrt[]{-3}}{2}$

The real zeros $=\frac{5}{2}$.