How do you factor $16-1/4k^2$ ?

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Answer 1 · 2016-09-23T20:00:36

$16-1/4k^2 = (4-1/2k)(4+1/2k)$

The difference of squares identity can be written:

$a^2-b^2 = (a-b)(a+b)$

Note that $16=4^2$ and $1/4k^2 = (1/2k)^2$ are both perfect squares.

So we find:

$16 - 1/4k^2 = 4^2-(1/2k)^2 = (4-1/2k)(4+1/2k)$

How do you factor 16-1/4k^2 16-1/4k^2?