Chapter 4, Problem 86RE

To determine

To find: the complex zeros of each polynomial function $f(x)$ .

Answer to Problem 86RE

Therefore, the complex zeros of $f(x)$ are $-\frac{1}{\sqrt{3}}i,\frac{1}{\sqrt{3}}i,2,$ and -3.

The factored form of $f$ is $\left(x-\frac{1}{\sqrt{3}}i\right)\left(x+\frac{1}{\sqrt{3}}i\right)(x+3)(x-2)$

Explanation of Solution

Given:

  $f(x)=3x^4+3x^3-17x^2+x-6$

Calculation:

Step 1: Determine the number of complex zeros.

Since the degree of the given polynomial is $4,$ the function will have 4 complex zeros.

Step 2: Apply the Descartes' Rule of Signs to determine the number of positive and negativereal zeros. There are three variations in the sign of the coefficients of $f(x)$ , whereas only 1variation in the sign of $f(-x)$ . Thus, the number of positive real zeros may be 3 and the number of negative real zeros may be 1

Step 3: Use the rational zeros theorem to find the possible potential rational zeros. The possible potential rational zeros of the given function are $\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3}$

Use the synthetic division to test whether 2 is a zero of the given function.

  $-1\overline{)\begin{array}{l}\text{3 3 -17 1 -6}\\ \underset{\_}{\text{ 6 18 2 6}}\\ \text{3 9 1 3 0}\end{array}}$

Since $f(2)=0,2$ is a zero of the given polynomial. The depressed equation of $f(x)$ is $3x^3+9x^2+$

  $x+3=0$

Step 4: Determine the remaining zeros. Factor the depressed equation by grouping.

Factor $3x^2$ from the first two terms and 1 from the last two terms

  $3x^2(x+3)+1(x+3)=0$

Factor out the common term $x+3$

  $\left(3x^2+1\right)(x+3)=0$

Apply the zero-product property.

  $3x^2+1=0$

  $\text{ or }x+3=0$

  $3x^2=-1\text{ or }x=-3x=-\frac{1}{\sqrt{3}}i,x=\frac{1}{\sqrt{3}}i$

Therefore, the complex zeros of $f(x)$ are $-\frac{1}{\sqrt{3}}i,\frac{1}{\sqrt{3}}i,2,$ and -3.

By the factor theorem, if $f(c)=0,$ then $x-c$ will be a factor of $f(x)$ .

Write $f$ in factored form.

  $f(x)=3x^4+3x^3-17x^2+1x-6=\left(x-\frac{1}{\sqrt{3}}i\right)\left(x+\frac{1}{\sqrt{3}}i\right)(x+3)(x-2)$

The factored form of $f$ is $\left(x-\frac{1}{\sqrt{3}}i\right)\left(x+\frac{1}{\sqrt{3}}i\right)(x+3)(x-2)$

Conclusion:

Therefore, the complex zeros of $f(x)$ are $-\frac{1}{\sqrt{3}}i,\frac{1}{\sqrt{3}}i,2,$ and -3.

The factored form of $f$ is $\left(x-\frac{1}{\sqrt{3}}i\right)\left(x+\frac{1}{\sqrt{3}}i\right)(x+3)(x-2)$