Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 5 - Polynomials and Polynomial Functions - 5-4 Dividing Polynomials - Practice and Problem-Solving Exercises - Page 308: 20


NOT a factor

Work Step by Step

Equat$x-3,$ to zero then solve for $x,$: \begin{align*} x-3&=0 \\ x&=3 \end{align*} Substituting $x= 3 $ in the given expression, $ x^3+4x^2+x-6 ,$ results to \begin{align*} & (3)^3+4(3)^2+(3)-6 \\&= 27+4(9)+3-6 \\&= 27+36+3-6 \\&= 60 .\end{align*} Since the substitution above is NOT equal to zero, then the remainder is NOT zero. By the Factor Theorem, $ x-3 $ is not a factor of $x^3+4x^2+x-6$.
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