Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 8 - Polynomials and Factoring - 8-7 Factoring Special Cases - Practice and Problem-Solving Exercises - Page 528: 57


$a.\quad(2-3n)(2+3n)(4+9n^{2})$ $b.\quad 16$ and $81n^{4}$ are both squares of square terms $c.\quad$ (sample answer)$\quad 16x^{4}-1$

Work Step by Step

a. $16-81n^{4}=$ ...write as the difference of two squares. ...($a=4,\ b=9n^{2})$ $=(4-9n^{2})(4+9n^{2})$ ...the expression in the first parentheses can be written as a difference of two squares. ...($a=2,b=3n$) $=(2-3n)(2+3n)(4+9n^{2})$ b. $16$ and $81n^{4}$ are both squares of square terms. c. Take for example, $2x$ and 1, square them twice $[(2x)^{2}]^{2} =[4x^{2}]^{2}=16x^{4},$ $[1^{2}]^{2}=1$ $16x^{4}-1$ can be factored twice: $=(4x^{2}-1)(4x^{2}+1)$ $=(2x+1)(2x-1)(4x^{2}+1)$
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