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)#