Chapter 10, Problem 2CR

To determine

The zeros of $9x^4+33x^3-71x^2-57x-10$ in the complex number system.

Answer to Problem 2CR

Solution:

The zeros of $9x^4+33x^3-71x^2-57x-10$ are $x_1=2,x_2=-5,x_3=-\frac{1}{3}$with $-\frac{1}{3}$being a repeated zero of multiplicity $2$ in the complex number system.

Explanation of Solution

Given information:

The polynomial function $9x^4+33x^3-71x^2-57x-10$

The degree of $f\left(x\right)$ is $4$ ; therefore, $f\left(x\right)$ has four complex zeros.

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

  $f\left(x\right)=9x^4+33x^3-71x^2-57x-10$ and $f\left(-x\right)=9x^4-33x^3-71x^2+57x-10$

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

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

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

The factor of $10$, all possible values of $p$ are $\pm 1,\pm 2,,\pm 5,\pm 10,$

The factor of $9$, all possible values of $q$ are $\pm 1,\pm 3,\pm 9$

The potential rational zeros are $\pm 1,\pm \frac{1}{3},\pm \frac{1}{9}\pm 2,\pm \frac{2}{3},\pm \frac{2}{9},\pm 5,\pm \frac{5}{3},\pm \frac{5}{9},\pm 10,\pm \frac{10}{3},\pm \frac{10}{9}$ .

  $\begin{array}{ccc}r& \text{coefficient}\text{of}quotient& \text{remainder}\\ 1& 942-29-86& -96\\ -1& 924-9538& -48\\ \frac{1}{3}& 936-59-\frac{130}{3}& -\frac{260}{3}\\ -\frac{1}{3}& 930-81-30& 0\\ -2& 915-101259& -528\end{array}$

The above table shows $-\frac{1}{3}$ is the real zero of $f\left(x\right)$

Since $f\left(-\frac{1}{3}\right)=0$, then $-\frac{1}{3}$ is a zero; therefore, $x-\left(-\frac{1}{3}\right)$ is a factor of $f\left(x\right)$

By using fourth row in the above table, the depressed equation is $f_1\left(x\right)=9x^3+30x^2-81x-30$

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

  $f_1\left(x\right)=9x^3+30x^2-81x-30$ and $f_1\left(-x\right)=-9x^3+30x^2+81x-30$ .

The number of positive real zeros of $f_1\left(x\right)$ are $1$, since changes in the sign of the coefficient of $f_1\left(x\right)$ are $1$ .

The number of negative real zeros of $f_1\left(x\right)$ are $2$, since changes in the sign of the coefficient of $f_1\left(-x\right)$ are $2$ .

By using the rational zeros theorem to find the potential rational zeros of $f_1\left(x\right)=9x^3+30x^2-81x-30$,

The factor of $30$, all possible values of $p$ are $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15$

The factor of $9$, all possible values of $q$ are $\pm 1,\pm 3,\pm 9$

The potential rational zeros are $\pm 1,\pm \frac{1}{3},\pm \frac{1}{9},\pm \frac{2}{3},\pm \frac{2}{9},\pm \frac{5}{3},\pm \frac{5}{9},\pm 2,\pm 3,\pm 5,\pm \frac{10}{3},\pm \frac{10}{9},\pm 6,\pm 10,\pm 15$

  $\begin{array}{ccc}r& \text{coefficient}\text{of}quotient& \text{remainder}\\ 1& 939-42& -72\\ -1& 921-60& -90\\ \frac{1}{3}& 933-70& -\frac{160}{3}\\ -\frac{1}{3}& 927-90& 0\\ \frac{2}{3}& 924-65& -160\end{array}$

The above table shows $-\frac{1}{3}$ is the real zero of $f\left(x\right)$

Since, $f\left(-\frac{1}{3}\right)=0$, then $-\frac{1}{3}$ is a zero; therefore, $x-\left(-\frac{1}{3}\right)$ is a factor of $f\left(x\right)$ .

By using fourth row in the above table, the depressed equation is $f_1\left(x\right)=9x^2+27x-90$

Equating with zero $9x^2+27x-90=0$

Divide $9$ from both sides,

  $\Rightarrow x^2+3x-10=0$

  $\Rightarrow x^2+5x-2x-10=0$

  $\Rightarrow \left(x+5\right)\left(x-2\right)=0$

  $\Rightarrow x+5=0\text{or}x-2=0$

Therefore, the remaining zeros are $x=-5$ and $x=2$ .

Thus, the complex zeros of $9x^4+33x^3-71x^2-57x-10$ are $x_1=2,x_2=-5,x_3=-\frac{1}{3}$ with $-\frac{1}{3}$ being a repeated zero of multiplicity $2$