Chapter 4.1, Problem 10CFU

To determine

To write: a polynomial equation of least degree for $6,2i,-2i,i,-i$ . Also find the degree and times graph of the related function cross the x-axis.

  $6,2i,-2i,i,-i$

Answer to Problem 10CFU

A polynomial equation with roots $6,2i,-2i,i\text{ and }-i\text{ is }{x}^5-6x^4+5x^3-30x^2+4x-24=0$

Since 4 of the roots are imaginary, the graph will only cross the x-axis once

Explanation of Solution

Given information: $6,2i,-2i,i,-i$

Calculation:

If $x=6,\text{ then }x-6$ is a factor of the polynomial.

Likewise, if $x=2i\text{ and }x=-2i,$ then $x-2i\text{ and }x+2i$ are factors of the polynomial

Same as this if $x=i\text{ and }x=-i,\text{ then }x-i\text{ and }x+i$ are factors of the polynomial.

Therefore, the linear factors for the polynomial are

  $\begin{array}{l}x-6\\ x-2i\\ x+2i\\ x-i\\ x+i\end{array}$

Now find the products of these factors.

  $\left(x-6\right)\left(x-2i\right)\left(x+2i\right)\left(x-i\right)\left(x+i\right)=0$

Multiply the factors

  $\begin{array}{l}\left(x-6\right)\left(x-2i\right)\left(x+2i\right)\left(x^2-i^2\right)=0\\ \left(x-6\right)\left(x-2i\right)\left(x+2i\right)\left(x^2+1\right)=0\\ \left(x-6\right)\left(x^2-4i^2\right)\left(x^2+1\right)=0\\ \left(x-6\right)\left(x^2+4\right)\left(x^2+1\right)=0\\ \left(x-6\right)\left(x^4+5x^2+4\right)=0\\ x^5-6x^4+5x^3-30x^2+4x-24=0\end{array}$

A polynomial equation with roots $6,2i,-2i,i\text{ and }-i\text{ is }{x}^5-6x^4+5x^3-30x^2+4x-24=0$

The degree of this equation is 5. Thus, the equation has an odd degree since 5 is an odd number.

Since 4 of the roots are imaginary, the graph will only cross the x-axis once.