Chapter 4.2, Problem 64AYU

In Problems 75-84, find the real solutions of each equation.

$2x^3+3x^2+2x+3=0$

To determine

To find: The real solutions of $f$.

Answer to Problem 64AYU

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

Explanation of Solution

Given:

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

$\text{F}$ 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

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

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

$-\text{ To }+ + \text{to} --\text{ to }+$

There will be 3 or 1 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 $\text{f}$,then $\text{p}$ is a factor of ${\text{a}}_{\text{0}}$ and $\text{q}$ is the factor of ${\text{a}}_{\text{n}}$.

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

Here ${\text{a}}_{\text{0}}\text{ = }3$ and ${\text{a}}_{\text{n}}\text{ = 2}$

Zeros of 2, $\text{q = }\pm 1, \pm 2$

Zeros of 2, $\text{p = }\pm 1, \pm 3$

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

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

Use long division or Synthetic division of $2x^3 + 3x^2 + 2x + 3$

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

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

$2x^3 + 3x^2 + 2x + 3 = 2\left(x + \frac{3}{2}\right)\left(x^2 + 1\right)$

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