How do you factor completely: $16x^2 – 9$ ?

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Answer 1 · 2015-07-11T05:52:59

$16x^2-9=(4x+3)(4x-3)$

$16x^2-9$

This is an example of the difference of squares, in which $a^2-b^2=(a+b)(a-b)$ .

$a=4x,$ $b=3$

$16x^2-9=(4x+3)(4x-3)$

How do you factor completely: 16x^2 – 916x^2 – 9?