College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Review Exercises - Page 75: 69


$(7m^4+3n)(7m^4-3n) $

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 49m^8-9n^2 ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Both $49m^8 \text{ and } 9n^2$ are perfect squares (the square root is exact) and separated by a minus sign. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} \\\\=(7m^4)^2-(3n)^2 \\\\=(7m^4+3n)(7m^4-3n) .\end{array}
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