Answer
$(4r-1)(5r+8)$
Work Step by Step
$20r^2+27r-8$
$20 \times -8=-160$
Factors of $-160$; Sum should $=27$
$32 \times -5$; Sum $=27$
$20r^2+32r-5r-8$
$=4r(5r+8)-1(5r+8)$
$=(4r-1)(5r+8)$
Algebra: A Combined Approach (4th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321726391
ISBN 13:
978-0-32172-639-1
Chapter 6 - Section 6.3 - Factoring Trinomials of the Form ax2+bx+c - Exercise Set - Page 435: 13
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