Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson

Chapter 3 - Polynomial and Rational Functions - Section 3.2 The Real Zeros of a Polynomial Function - 3.2 Assess Your Understanding - Page 224: 47

Answer

$f(x)=(2x-1)(x^2+1)$ Real Zeros: $\dfrac{1}{2}$ with multiplicity $1$.

Work Step by Step

Let us consider that $m$ is a factor of the constant term and $n$ is a factor of the leading coefficient. Then the potential zeros can be expressed by the possible combinations as: $\dfrac{m}{n}$. We see from the given polynomial function that it has at most $3$ real zeros as degree is $3$. The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1$and $n=\pm 1, \pm 2$ Therefore, the possible rational roots of $f(x)$ are: $\dfrac{m}{n}=\pm 1, \pm \dfrac{1}{2}$ We test with synthetic division; we will try $x-\dfrac{1}{2}$. $\left.\begin{array}{l} \dfrac{1}{2} \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 2 & -1 &2&-1\\\hline &1&0 &1\\\hline 2&0 &2& |\ \ 0\end{array}$ Thus, we have: $f(x)=(x-\dfrac{1}{2})(2x^2+2) \\=(2x-1)(x^2+1)$ Real Zeros: $\dfrac{1}{2}$ with multiplicity $1$.

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