Chapter 2.5, Problem 12E

To determine

To verify: The function $f\left(x\right)=x^3+9x$ has the indicated roots $0,-3i,3i$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=x^3+9x$ and the roots $0,-3i,3i$ .

Formula used:

Complex zeros of polynomial functions are imaginary numbers that cannot be plotted or represented on the graph.

To prove that the roots are the complex zeros of a polynomial function, then substitute the given roots in the function and the value of function should come out to be zero.

The value of the imaginary number iota (i) is $\begin{array}{l}i=\sqrt{-1}\\ i^2=-1\\ i^3=-i\end{array}$ .

Proof:

Consider the function, $f\left(x\right)=x^3+9x$ .

Substitute $x=0$ in the function $f\left(x\right)=x^3+9x$ as,

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

Now, substitute $x=-3i$ in the function $f\left(x\right)=x^3+9x$ as,

  $\begin{array}{c}f\left(x\right)=x^3+9x\\ f\left(-3i\right)={\left(-3i\right)}^3+9\left(-3i\right)\\ =\left(-27\right)\left(-i\right)+\left(-27\right)\left(i\right)\\ =0\end{array}$

Substitute $x=3i$ in the function $f\left(x\right)=x^3+9x$ as,

  $\begin{array}{c}f\left(x\right)=x^3+9x\\ f\left(3i\right)={\left(3i\right)}^3+9\left(3i\right)\\ =\left(27\right)\left(-i\right)+\left(27\right)\left(i\right)\\ =0\end{array}$

Since, all the roots give value of the function as zero; thus, it is verified that given roots $0,-3i,3i$ are the complex zeros of the polynomial function $f\left(x\right)=x^3+9x$ .