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

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Answer 1 · 2016-05-13T13:10:14

A very slightly different way of presenting the calculation:

$" "(3x-4)(3x+4)$

Write as: $" "3^2x^2-4^2$

Which is the same as: $" "(3x)^2-4^2$

It is known that standard form of $a^2-b^2=(a+b)(a-b)$

So we have: $" "(3x-4)(3x+4)$

How do you factor completely 9x^2 - 169x^2 - 16?