Answer
$(4x+\frac{1}{3})(4x-\frac{1}{3})$
Work Step by Step
$16x^2-\frac{1}{9}=(4x)^2-(\frac{1}{3})^2=(4x+\frac{1}{3})(4x-\frac{1}{3})$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Prerequisites - p.4 - Factoring Polynomials - Exercises - Page 39: 71
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