Chapter 4.1, Problem 36E

To determine

To write:a polynomial equation of least degree for $-5i,-i,i,5i$ and find the degree as even or odd. Also find the times graph of the related function cross the x-axis.

Answer to Problem 36E

  $x^4+26x^2+25=0$

The degree of this equation is 4, even degree

Since all the roots are imaginary, the graph will not cross the x-axis

Explanation of Solution

Given information: $-5i,-i,i,5i$

Calculation:

If $x=-5i,$ then

  $x+5i\text{ is }$ a factor of the polynomial.

If $x=-5i,$ then $x+5i$ is a factor of the polynomial.

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

Therefore, the linear factors for the polynomial are

  $\begin{array}{l}\left(x+5i\right)\\ \left(x-5i\right)\\ \left(x+i\right)\\ \left(x-i\right)\end{array}$

Now find the product of these factors.

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

Multiply the factors

  $\begin{array}{l}\left(x+5i\right)\left(x-5i\right)\left(x+i\right)\left(x-i\right)=0\\ \left(x^2-25i^2\right)\left(x^2-i^2\right)=0\\ \left(x^2+25\right)\left(x^2+1\right)=0\\ x^4+x^2+25x^2+25=0\\ x^4+26x^2+25=0\end{array}$

A polynomial equation with roots $-5i,-i,i,5i$ is

  $x^4+26x^2+25=0$

The degree of this equation is 4. Thus the equation has an evendegree since 4 is an even number.

Since all the roots are imaginary, the graph will not cross the x-axis