Chapter 4.3, Problem 23AYU

In Problems 23-30, use the given zero to find the remaining zeros of each function.

$f\left(x\right)=x^3-4x^2+4x-16$; zero: $2i$

To determine

To find: The remaining zeros of the polynomial function $f\left(x\right)$ whose coefficients are real numbers.

Answer to Problem 23AYU

$\text{Factored form off }f(x)=x^3-4x^2+4x-16=(x-4)(x-2i)(x+2i)$.

Explanation of Solution

Given:

$f\left(x\right)=x^3-4x^2+4x-16$

$\text{Zero }=2i$

Degree of the polynomial function $f\left(x\right)=x^3-4x^2+4x-16=3$.

Therefore $f\left(x\right)$ have 3 zeros.

From Descartes’ rule of signs.

$f\left(x\right)=x^3-4x^2+4x-16$

$+\text{ to }--\text{ to }+$

Therefore $f\left(x\right)=x^3-4x^2+4x-16$ have 2 positive real zeros or no positive real zeros.

Rational zeros theorem provides information about the potential rational zeros of the polynomial function with integer coefficients.

Zeros of $a_0=p=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16$.

Zeros of $a_n=q=\pm 1$.

By Synthetic division.

The coefficients of $f\left(x\right)=1 -4 4 -16$.

The positive real zero $=4$.

The depressed equation $=x^2+4$.

The depressed equation can be factored further $=x^2=-4x=\text{ ±}2i$.

The complex zeros $=\pm 2i$.

The real zeros $=4$.

Factored form of $f\left(x\right)=x^3-4x^2+4x-16=\left(x-4\right)\left(x–2\text{i}\right)\left(x+2\text{i}\right)$.