Chapter 5.5, Problem 48PPSE

Solution

(a)

To calculate : All possible combinations of root type using real and imaginary as types of roots.

All the four roots of the polynomial $f\left(x\right)$ are real roots. The polynomial $f\left(x\right)$ has two real roots and two imaginary roots. All the four roots of the polynomial $f\left(x\right)$ are imaginary roots.

Given information :

A fourth-degree polynomial function.

Calculation :

Let $f\left(x\right)$ be a fourth degree polynomial such that function is $f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e$ .

Determine all the possible combinations of root of the above polynomial.

A fourth degree polynomial has four roots. So, there are three possible combinations of roots of the polynomial $f\left(x\right)$ .

1. All the four roots of the polynomial $f\left(x\right)$ are real roots.
2. The polynomial $f\left(x\right)$ has two real roots and two imaginary roots. This is because, according to Conjugate Root Theorem, complex conjugate root always exits in pair. Hence, imaginary roots will always be in pair.
3. All the four roots of the polynomial $f\left(x\right)$ are imaginary roots because imaginary roots always come in pairs.

(b)

To calculate : All possible combinations of root type using real and imaginary as types of roots.

All the five roots of the polynomial $g\left(x\right)$ are real roots. The polynomial $g\left(x\right)$ has one real root and four imaginary roots. There can be three real roots and two imaginary of the polynomial $g\left(x\right)$ .

Given information :

A fifth-degree polynomial function.

Calculation :

Let $g\left(x\right)$ be a fourth degree polynomial such that function is $g\left(x\right)=l{x}^5+m{x}^4+n{x}^3+p{x}^2+qx+k$ .

Determine all the possible combinations of root of the above polynomial.

A fifth degree polynomial has five roots. So, there are three possible combinations of roots of the polynomial $g\left(x\right)$ .

1. All the five roots of the polynomial $g\left(x\right)$ are real roots.
2. The polynomial $g\left(x\right)$ has one real root and four imaginary roots. This is because, according to Conjugate Root Theorem, complex conjugate root always exits in pair. Hence, imaginary roots will always be in pair.
3. There can be three real roots and two imaginary of the polynomial $g\left(x\right)$ since imaginary roots always come in pairs.

(c)

To determine : A conjecture about the number of real roots.

Every odd degree polynomial $h\left(x\right)$ has at least one real root.

Given information :

An odd-degree polynomial equation.

Calculation :

Let $h\left(x\right)$ be an odd degree polynomial such that function is $h\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+\mathrm{...}+\mathrm{...}+a_{1}x+a_0$ .

An odd degree polynomial has always odd number of roots.

The, there are two possible cases

1. All the roots of the polynomial $h\left(x\right)$ are real roots.
2. The polynomial $h\left(x\right)$ has at least one real root and the other $n-1$ in every number is imaginary roots. This is because, according to Conjugate Root Theorem, complex conjugate root always exits in pair. Hence, imaginary roots will always be in pair.

Hence, every odd degree polynomial $h\left(x\right)$ has at least one real root.