Chapter 4, Problem 8CT

In Problems $8$ and $9$, write a function that meets the given condition.

Fourth – degree polynomial with real coefficients; zeros; $-2,0,3+i$

To determine

The fourth-degree polynomial function with real coefficient using the information zeros: $-2,0,3+i$.

Answer to Problem 8CT

Solution:

The fourth-degree polynomial with zeros $-2,0,3+i$ is $f\left(x\right)=a\left(x^4-4x^3-2x^2+20x\right)$.

Explanation of Solution

Given Information:

Degree 4; zeros: $-2,0,3+i$.

Let $f$ be the polynomial function with real coefficient of degree 4 whose zeros are $x=-2,x=0,x=3+i$.

Since degree of polynomial is 4, there must be 4 zeros for the polynomial $f$.

By using Conjugate Pairs Theorem, if $x=a+bi$ is a zero of $f$, then the complex conjugate $x=a-bi$ is also zero of $f$.

Since $3+i$ is a zero, by the Conjugate Pairs Theorem $3-i$ is also a zero of $f$.

Therefore the all zeros of polynomial function $f$ are $-2,0,3+i$ and $3-i$.

By Factor Theorem, if $f\left(c\right)=0$ then $x-c$ is a factor of $f\left(x\right)$.

So, $f\left(x\right)$ can be written as

$f\left(x\right)=a\left(x-0\right)\left(x+2\right)\left[x-\left(3+i\right)\right]\left[x-\left(3-i\right)\right]$, where $a$ is any nonzero real number.

Simplify this,

$=ax\left(x+2\right)\left[x-\left(3+i\right)\right]\left[x-\left(3-i\right)\right]$

$=a\left(x^2+2x\right)\left[\left(x-3\right)-i\right]\left[\left(x-3\right)+i\right]$

$=a\left(x^2+2x\right)\left[{\left(x-3\right)}^2-i^2\right]$

$=a\left(x^2+2x\right)\left[x^2-6x+9-\left(-1\right)\right]$

$=a\left(x^2+2x\right)\left(x^2-6x+10\right)$

$=a\left(x^4-4x^3-2x^2+20x\right)$

Therefore, $f\left(x\right)=a\left(x^4-4x^3-2x^2+20x\right)$ is fourth-degree polynomial with zeros: $-2,0,3+i$.