How do you factor completely #169x^2 – 64#?

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Answer 1 · 2017-04-17T13:05:29

See the entire solution process below:

This problem is a special form of this rule:

#(a + b)(a + b) = a^2 - b^2# or #a^2 - b^2 = (a + b)(a - b)#

If we let #a = 13x# and let #b = 8# and substitute we get:

#169x^2 - 64 = (13x + 8)(13x - 8)#

Answer 2 · 2017-04-17T13:07:27

You may notice that both #169and64# are squares.

#=13^2*x^2-8^2#

#=(13x)^2-8^2#

We now use the special product #A^2-B^2harr(A+B)(A-B)#

#=(13x+8)(13x-8)#