Chapter 2.5, Problem 62E

To determine

Tofind:Find the zero’s of the polynomial function.

Answer to Problem 62E

The zeros of the polynomial function are $x=-1,3i,-3i$

Explanation of Solution

Given information: The given polynomial function is $f\left(x\right)=x^3+x^2+9x+9$ and one root is $3i$

From fundamental theorem of algebra we know that every polynomial of degree n has exactly n roots.

So, the given polynomial has exactly 3 roots.

We have the polynomial function is $f\left(x\right)=x^3+x^2+9x+9$ and one imaginary root is $3i$

We know that imaginary roots comes in pair.

i.e if $\alpha +i\beta$ then $\alpha -i\beta$ must be a root of the polynomial function.

So, $x=-3i$ must be a root.

Now, for the third root we put

  $\begin{array}{l}f\left(x\right)=0\\ x^3+x^2+9x+9=0\\ x^2\left(x+1\right)+9\left(x+1\right)=0\\ \left(x+1\right)\left(x^2+9\right)=0\end{array}$

This give $x^2+9=0$ and $x+1=0$

  $x^2+9=0\Rightarrow x=\pm 3i$

  $x+1=0\Rightarrow x=-1$

Hence, the roots of the polynomial are $x=-1,3i,-3i$