   # Let f ( x ) = a 0 + a 1 x + ⋯ + a n x n in ℝ [ x ] . Prove that if a i ≥ 0 for all i = 0 , 1 , … , n or if a i ≤ 0 for all i = 0 , 1 , … , n , then f ( x ) has no positive zeros. ### Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
Publisher: Cengage Learning,
ISBN: 9781285463230

#### Solutions

Chapter
Section ### Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
Publisher: Cengage Learning,
ISBN: 9781285463230
Chapter 8.4, Problem 30E
Textbook Problem
1 views

## Let f ( x ) = a 0 + a 1 x + ⋯ + a n x n in ℝ [ x ] . Prove that if a i ≥ 0 for all i = 0 , 1 , … , n or if a i ≤ 0   for all   i = 0 , 1 , … , n , then f ( x ) has no positive zeros.

To determine

To prove: If ai0 for all i=0,1,2,3,...,n or if ai0 for all i=0,1,2,3,...,n, then f(x) has no positive zeros.

### Explanation of Solution

Given information:

f(x)=a0+a1x+...+anxn in [x].

Formula used:

Let f(x) be a polynomial over the field F. If c is an element of F, such that f(c)=0, then c is called a zero of f(x).

Proof:

Let f(x)=a0+a1x+...+anxn in [x], where is the field of real numbers.

Let ai0 for all i=0,1,2,3,...,n.

Assume that α is a positive zero of f(x).

Implies that f(α)=0

Then, 0=a0+a1α+...+anαn

As α>0,ai0

implies that each ai must be zero

That is, f(x)=0 means α=0.

This is not true.

Hence, there is a contraction.

Thus, α is not a positive zero of f(x)

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