## Algebra 1: Common Core (15th Edition)

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