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 $A^2-B^2harr(A+B)(A-B)$

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

How do you factor completely 169x^2 – 64169x^2 – 64?