College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 3, Polynomial and Rational Functions - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 320: 80

Answer

$1$ is an upper bound for the real zeros of $P$ $-1$ is a lower bound for the real zeros of $P$

Work Step by Step

Upper and Lower bound Theorem states that: Suppose $f$ is a polynomial of degree $n\geq 1$. If $c>0$ is synthetically divided into $f$ and all of the coefficients of the quotient and remainder are all non-negative values, then $c$ is an upper bound for the real zeros of $f$. That is, there are no real zeros greater than $c$. If $c<0$ is synthetically divided into $f$ and the coefficients of the quotient and remainder alternate signs, then $c$ is a lower bound for the real zeros of $f$. That is, there are no real zeros less than $c$. If the number $0$ occurs as a coefficient of the quotient or remainder in the final line of the division table in either of the above cases, it can be treated as (+) or (−) as needed. $P(x)=x^5-x^4+1$; let's try $1$: $\begin{array}{lllll} \underline{1}| & 1&-1 & 0 & 0 & 0& 1 \\ & & 1& 0& 0 & 0&0 \\ \hline & & & & \\ & 1&0 & 0 & 0 & 0&1 \end{array}$ All of the coefficients of the quotient and remainder are all non-negative values. thus, $1$ is an upper bound for the real zeros of $P$. Let's try $-1$: $\begin{array}{lllll} \underline{-1}| & 1&-1& 0 & 0 & 0& 1\\ & &-1& 2&-2 & 2&-2\\ \hline & & & & \\ & 1&-2& 2& -2 & 2&-1 \end{array}$ The coefficients of the quotient and remainder alternate signs, thus, $-1$ is a lower bound for the real zeros of $P$.
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