## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Section 2.4 - Dividing Polynomials; Remainder and Factor Theorems - Exercise Set - Page 364: 49

#### Answer

$x=\frac{1}{3},\frac{1}{2},1$

#### Work Step by Step

Step 1. With the given table, we can determine a solution of the equation as $x=1$. Step 2. The coefficients of the dividend $f(x)=6x^3-11x^2+6x-1$ can be identified in order as $\{6,-11,6,-1\}$ and the divisor is $x-1$; use synthetic division as shown in the figure to get the quotient and the remainder. Step 3. We can identify the result as $f(x)=(6x^2-5x+1)(x-1)=(2x-1)(3x-1)(x-1)$; thus we can find the zeros of $f(x)$ as $x=\frac{1}{3},\frac{1}{2},1$

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