Answer
$P(2)=Q(2)=5$
$R(x)=((((x-2)x+3)x-2)x+3)x+4$
$R(3)=157$
Work Step by Step
$Q(x)=(((3x-5)x+1)-3)x+5
=((3x^2-5x+1)x-3)x+5
=(3x^3-5x^2+x-3)x+5
=3x^4-5x^3+x^2-3x+5$
$P(2)=Q(2)=5$
$R(x)=((((x-2)x+3)x-2)x+3)x+4$
$R(3)=157$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 3 - Section 3.3 - Dividng Polynomials - 3.3 Exercises - Page 275: 76
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