Chapter 4.2, Problem 65AYU

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

$3x^3+4x^2-7x+2=0$

To determine

To find: The real solutions of $f$.

Answer to Problem 65AYU

$\text{The real solution = }\frac{2}{3}, -1 \pm \sqrt{2}$.

Explanation of Solution

Given:

$f\left(x\right) = 3x^3 + 4x^2 - 7x + 2 = 0$

$\text{F}$ 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

$f\left(x\right) = 3x^3 + 4x^2 - 7x + 2$

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

There will be 2 or 0 positive real zeros.

$f\left(-x\right) = -3x^3 + 4x^2 + 7x + 2$

$-\text{ To }+$

There will be 1 negative real zero.

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 ${\text{a}}_{\text{n}}$.

$f\left(x\right) = 3x^3 + 4x^2 - 7x + 2$

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

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

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

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

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

Use long division or Synthetic division of $3x^3 + 4x^2 - 7x + 2$

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

The depressed equation $= 3\left(x^2 + 2x - 1\right)$

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

Factorized further

$\text{a = 1, b = 2}$ and $\text{c = }-1$

$x = \frac{-b \pm \sqrt{b^2 - \mathrm{4 . }a . c}}{\mathrm{2 . }a} = \frac{-2 \pm \sqrt{2^2 - \mathrm{4 . 1 . }\left(-1\right)}}{\mathrm{2 . 1}} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2}$

$x = -1 \pm \sqrt{2}$

The real zeros $= \frac{2}{3}, -1 \pm \sqrt{2}$.