Chapter 2.5, Problem 13E

To determine

To verify: The function $f\left(x\right)=3x^4-48$ has the indicated roots $-2,2,-2i,2i$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=3x^4-48$ and the roots $-2,2,-2i,2i$ .

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, $f\left(x\right)=3x^4-48$ .

Substitute $x=-2$ in the function $f\left(x\right)=3x^4-48$ as,

  $\begin{array}{c}f\left(x\right)=3x^4-48\\ f\left(-2\right)=3{\left(-2\right)}^4-48\\ =48-48\\ =0\end{array}$

Now, substitute $x=2$ in the function $f\left(x\right)=3x^4-48$ as,

  $\begin{array}{c}f\left(x\right)=3x^4-48\\ f\left(2\right)=3{\left(2\right)}^4-48\\ =48-48\\ =0\end{array}$

Substitute $x=-2i$ in the function $f\left(x\right)=3x^4-48$ as,

  $\begin{array}{c}f\left(x\right)=3x^4-48\\ f\left(-2i\right)=3{\left(-2i\right)}^4-48\\ =48-48\\ =0\end{array}$

Now, substitute $x=2i$ in the function $f\left(x\right)=3x^4-48$ as,

  $\begin{array}{c}f\left(x\right)=3x^4-48\\ f\left(2i\right)=3{\left(2i\right)}^4-48\\ =48-48\\ =0\end{array}$

Since, all the roots give value of the function as zero; thus, it is verified that given roots $-2,2,-2i,2i$ are the complex zeros of the polynomial function $f\left(x\right)=3x^4-48$ .