Answer
$(5x+8)(4x-1)$
Work Step by Step
$20x^{2}+27x-8=$
two factors of $ac=-160$ whose sum is $b=+27$ are $+32$ and $-5.$
Rewrite the middle term and factor in pairs.
$20x^{2}+27x-8=20x^{2}+32x-5x-8$
$=4x(5x+8)-(5x+8)$
$=(5x+8)(4x-1)$
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