Chapter 4, Problem 25RE

To determine

To calculate: The remaining zeros of f and a polynomial function with real coefficients that has the zeros.

Answer to Problem 25RE

The remaining zeros are $4-i$ .

The polynomial function is $x^3-14x^2+65x-102$ .

Explanation of Solution

Given information:

Degree 3; zeros: $4+i,6$ .

Formula used:

Conjugate Pairs Theorem:

The theorem states that if the coefficients of the polynomial $f(x)$ are real numbers, whose zeros are $r=a+bi$ then the complex conjugate $\overline{r}=a-bi$ is also a zero of f .

Linear Factorization Theorem:

If $r_1,r_2,\mathrm{....},r_n$ are zeros of polynomial function $f(x)$ , then $f(x)$ can be written as

  $f(x)=a(x-r_1)(x-r_2)\mathrm{.......}(x-r_n)$ where $a$ is a nonzero constant.

Calculation:

Consider the zeros: $4+i,6$

Since coefficients of f are real numbers, complex zeros appear as conjugate pairs.

This implies $4-i$ is also a zeros of f .

Hence, the remaining zeros of f are $4+i,6,4-i$ .

Using Linear Factorization Theorem $f(x)$ can be written as :

  $f(x)=a(x-6)(x-(4+i))(x-(4-i))$

Here , $a=1$ .

  $\begin{array}{l}f(x)=a(x-6)(x-(4+i))(x-(4-i))\\ =(x-6)\left[x^2-x(4-i)-x(4+i)+(4+i)(4-i)\right]\\ =(x-6)\left[x^2-4x+ix-4x-xi+\left({(4)}^2-i^2\right)\right]\end{array}$

  $\begin{array}{l}=(x-6)\left[x^2-8x+\left(16-i^2\right)\right]\\ =(x-6)\left[x^2-8x+\left(16-(-1)\right)\right]\\ =(x-6)\left[x^2-8x+\left(16+1\right)\right]\\ =(x-6)\left[x^2-8x+17\right]\end{array}$ where $i^2=-1$

  $\begin{array}{l}=x\left[x^2-8x+17\right]-6\left[x^2-8x+17\right]\\ =x^3-8x^2+17x-6x^2+48x-102\\ =x^3-14x^2+65x-102\end{array}$

Therefore , the polynomial function is $x^3-14x^2+65x-102$ .