Chapter 4, Problem 84RE

To determine

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

Answer to Problem 84RE

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

The factored form of $f$ is $(x-\sqrt{2i})(x+\sqrt{2i}){(x+3)}^2$ .

Explanation of Solution

Given:

  $f(x)=x^4+6x^3+11x^2+12x+18$

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 is no variation in the sign of the coefficients of $f(x)$ , whereas 4 variations inthe sign of $f(-x)$ . Thus, the number of positive real zeros may be 0 and number of negative realzeros may be 4,2 or 0

Step 3: Use the rational zeros theorem to find the possible potential rational zeros. Thepossible potential rational zeros of the given function are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$

Use the synthetic division to test whether -1 is a zero of the given function.

  $-1\overline{)\begin{array}{l}1\text{ 6 11 12 18}\\ \underset{\_}{\text{ -1 -5 -6 -6}}\\ 1\text{ 5 6 6 12}\end{array}}$

We note that -1 is not a zero of the given polynomial.

Now, test with another possible zero, say, -3.

  $-1\overline{)\begin{array}{l}1\text{ 6 11 12 18}\\ \underset{\_}{\text{ -3 -9 -6 -18}}\\ 1\text{ 3 2 6 0}\end{array}}$

Since $f(-3)=0,-3$ is a zero of the given polynomial. The depressed equationof $f(x)$ is $x^3+3x^2+2x+6=0$

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

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

  $x^2(x+3)+2(x+3)=0$

Factor out the common term $x+3$

  $\left(x^2+2\right)(x+3)=0$

Apply the zero-product property.

  $x^2+2=0$

  $\text{ or }x+3=0$

  $x^2=-2$

  $\text{ or }x=-3$

  $x=-\sqrt{2i},x=\sqrt{2i}$ or $x=-3$

The zero-3 has a multiplicity 2

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

  $f(x)=x^4+6x^3+11x^2+12x+18=(x-\sqrt{2i})(x+\sqrt{2i}){(x+3)}^2$

The factored form of $f$ is $(x-\sqrt{2i})(x+\sqrt{2i}){(x+3)}^2$

Conclusion:

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

The factored form of $f$ is $(x-\sqrt{2i})(x+\sqrt{2i}){(x+3)}^2$