How do you factor completely $49x^2-9$ ?

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Answer 1 · 2016-06-29T17:09:36

$=(7x)^2-3^2=(7x-3)(7x+3).$

We know that, $A^2-B^2=(A-B)(A+B).$

Given Exp. $=(7x)^2-3^2=(7x-3)(7x+3).$

Answer 2 · 2016-06-29T17:10:16

$49x^2-9=(7x+3)(7x-3)$

As both the monomials in the binomial are squares, with a negative sign between them, we can the identity $a^2-b^2=(a+b)(a-b)$

Here we have $49x^2-9$

= $(7x)^2-3^2$

= $(7x+3)(7x-3)$

How do you factor completely 49x^2-949x^2-9?