Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 10 - Section 10.5 - Finding Factors of Special Products - Exercise - Page 354: 14

Answer

$$4(2x+5)(2x-5)$$

Work Step by Step

$16x^{2}-100$ In this case we can factor out a 4, so $4(4x^{2}-25)$ This is the difference between two squares. There is a simple formula for this case: $x^{2}-y^{2}=(x+y)(x-y)$ So, applying this formula, We need to take the square of both terms. $\sqrt 4x^{2}=2x$ $\sqrt 25=5$ So the answer would look like: $$4(2x+5)(2x-5)$$ (One with a positive sign and one with a negative sign and remember to multiply it by the common factor you took out!)
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