Chapter 5.6, Problem 47PPSE

Solution

To show: At least one real root is possible for any odd degree polynomial equation with real coefficient using Fundamental Theorem of algebra.

Given information: It is given that any odd degree polynomial equation with real coefficient is present.

Concept Used:

Fundamental theorem of algebra:

A polynomial function with degree $n$ will have $n$ zeros, they can be Real or Non-real

This Implies that a polynomial function can have at most $n$ real zeros

Conjugate Root Theorem:

If polynomial equation with real coefficients has complex roots

Then the Complex roots exist in Conjugate pairs

That is, if $a+ib$ is a root, then $a-ib$ must be a root

Explanation:

A polynomial equation with odd degree will have odd number of zeros

Some of them will be Real and some of them will be complex

But the polynomial equation is having real coefficients, therefore there should be even number of complex roots.

It is because by Conjugate root theorem, when the coefficients are all real, the polynomial will have its complex roots in conjugate pairs.

So there are odd number of roots, out of which some of them may be complex

Therefore, there are odd number of real roots.

  $\text{Total number of roots - Number of complex roots = Number of real roots}$

Because there are odd number of real roots, there should be more than one real root.

Because 1 is the smallest odd number.