Chapter 6.2, Problem 48E

To determine

To find: the polynomial of least degree with the given roots.

Answer to Problem 48E

  $x^3+4x^2+9x+36=0$

Explanation of Solution

Given information:

The roots are $-4,3i$ and $-3i$ .

Calculation:

The factors for given roots will be as follows:

  $\begin{array}{ccc}\begin{array}{l}\text{Root}\\ x=-4\\ x=3i\\ x=-3i\end{array}& \begin{array}{l}\\ \\ \\ \end{array}& \begin{array}{l}\text{Factor of polynomial }\\ x+4\\ x+3i\\ x-3i\end{array}\end{array}$

Now product these factors as follows:

  $\begin{array}{c}\left(x+4\right)\left(x+3i\right)\left(x-3i\right)=0\\ \left(x+4\right)\left(x^2-{\left(3i\right)}^2\right)=0\\ \left(x+4\right)\left(x^2-9i^2\right)=0\\ \left(x+4\right)\left(x^2+9\right)=0\\ x^3+4x^2+9x+36=0\end{array}$

Thus the polynomial equation is $x^3+4x^2+9x+36=0$ .