Chapter 2.5, Problem 14E

To determine

To verify: The function $f\left(x\right)=2x^5-2x$ has the indicated roots $0,1,-1,-i,i$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=2x^5-2x$ and the roots $0,1,-1,-i,i$ .

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

  $\begin{array}{l}i=\sqrt{-1}\\ i^2=-1\\ i^3=-i\\ i^4=1\end{array}$

Proof:

Consider the function, $f\left(x\right)=2x^5-2x$ .

Substitute $x=0$ in the function $f\left(x\right)=2x^5-2x$ as,

  $\begin{array}{c}f\left(x\right)=2x^5-2x\\ f\left(0\right)=2{\left(0\right)}^5-2\left(0\right)\\ =0-0\\ =0\end{array}$

Now, substitute $x=1$ in the function $f\left(x\right)=2x^5-2x$ as,

  $\begin{array}{c}f\left(x\right)=2x^5-2x\\ f\left(1\right)=2{\left(1\right)}^5-2\left(1\right)\\ =2-2\\ =0\end{array}$

Substitute $x=-1$ in the function $f\left(x\right)=2x^5-2x$ as,

  $\begin{array}{c}f\left(x\right)=2x^5-2x\\ f\left(-1\right)=2{\left(-1\right)}^5-2\left(-1\right)\\ =-2+2\\ =0\end{array}$

Substitute $x=-i$ in the function $f\left(x\right)=2x^5-2x$ as,

  $\begin{array}{c}f\left(x\right)=2x^5-2x\\ f\left(-i\right)=2{\left(-i\right)}^5-2\left(-i\right)\\ =-2i+2i\\ =0\end{array}$

Now, substitute $x=i$ in the function $f\left(x\right)=2x^5-2x$ as,

  $\begin{array}{c}f\left(x\right)=2x^5-2x\\ f\left(i\right)=2{\left(i\right)}^5-2\left(i\right)\\ =2i-2i\\ =0\end{array}$

Since, all the roots give value of the function as zero; thus, it is verified that given roots $0,-1,1,-i,i$ are the complex zeros of the polynomial function $f\left(x\right)=2x^5-2x$ .