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 309: 34

Answer

$P(a)=0$

Work Step by Step

Using the synthetic division shown below, the remainder (bottom-right number) of $( x^3-7x^2+15x-9 )\div( x-3 ),$ is \begin{align*} 0 .\end{align*} Substituting $x= 3 $ in $P(x)= x^3-7x^2+15x-9 ,$ then by the Remainder Theorem, the remainder when $P(x)$ is divided by $ x-3 $ is \begin{align*} P(3)&= (3)^3-7(3)^2+15(3)-9 \\&= 27-7(9)+15(3)-9 \\&= 27-63+45-9 \\&= 0 .\end{align*} The two solutions above show that the remainder, $P(a),$ is \begin{align*} P(a)=0 .\end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.