Chapter 4, Problem 4CT

To determine

The zeros of $3x^3+2x-1=8x^2-4$ in the complex number system.

Answer to Problem 4CT

Solution:

The complex zeros of $3x^3-8x^2+2x+3=0$ are $1,\frac{5+\sqrt{61}}{6}$ and $\frac{5-\sqrt{61}}{6}$ .

Explanation of Solution

Given information:

The polynomial function, $3x^3+2x-1=8x^2-4$

Let the equation $3x^3+2x-1=8x^2-4$ .

Subtract $8x^2$ on both sides,

  $3x^3-8x^2+2x-1=-4$

Add $4$ on both sides,

  $3x^3-8x^2+2x+3=0$

Therefore, $f\left(x\right)=3x^3-8x^2+2x+3$ .

The degree of $f\left(x\right)$ is $3$ therefore, $f\left(x\right)$ have three complex zeros.

By using the Descartes Rule of signs to find the real zeros,

  $f\left(x\right)=3x^3-8x^2+2x+3$ and $f\left(-x\right)=-3x^3-8x^2-2x+3$ .

The number of positive real zeros of $f\left(x\right)$ are $2$ , since changes in the sign of the coefficient of $f\left(x\right)$ are $3$ .

The number of negative real zeros of $f\left(x\right)$ are $1$ , since changes in the sign of the coefficient of $f\left(-x\right)$ are $1$ .

By using the Rational zeros Theorem to find the potential rational zeros of $f\left(x\right)$ ,

The factor of $3$ , all possible value of $p$ are $\pm 1,\pm 3$ .

The factor of $3$ , all possible value of $q$ are $\pm 1,\pm 3$ .

The potential rational zeros are $\pm 1,\pm 3,\pm \frac{1}{3}$ .

  $\begin{array}{ccc}r& \text{coefficient}\text{of}q\left(x\right)& \text{remainder}\\ -1& 3-1113& -10\\ 1& 3-5-3& 0\end{array}$

Since, $f\left(1\right)=0$ then $1$ is a zero therefore, $x-1$ is a factor of $f\left(x\right)$ .

Then, the depressed equation is, $3x^2-5x-3=0$ .

By quadratic formula; $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

  $\Rightarrow x=\frac{5\pm \sqrt{{(-5)}^2-4(3)(-3)}}{6}$

  $\Rightarrow x=\frac{5\pm \sqrt{61}}{6}$

Therefore, $x=\frac{5+\sqrt{61}}{6}$ and $x=\frac{5-\sqrt{61}}{6}$ .

Therefore, the three zeros of $f$ are $1,\frac{5+\sqrt{61}}{6}$ and $\frac{5-\sqrt{61}}{6}$ .