## Algebra: A Combined Approach (4th Edition)

$(4r-1)(5r+8)$
$20r^2+27r-8$ $20 \times -8=-160$ Factors of $-160$; Sum should $=27$ $32 \times -5$; Sum $=27$ $20r^2+32r-5r-8$ $=4r(5r+8)-1(5r+8)$ $=(4r-1)(5r+8)$