Question 590847

{{{x^2-64}}} Start with the given expression.

{{{(x)^2-64}}} Rewrite {{{x^2}}} as {{{(x)^2}}}.

{{{(x)^2-(8)^2}}} Rewrite {{{64}}} as {{{(8)^2}}}.

Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=x}}} and {{{B=8}}}.

So let's use the difference of squares formula {{{A^2-B^2=(A+B)(A-B)}}} to factor the expression:

{{{A^2-B^2=(A+B)(A-B)}}} Start with the difference of squares formula.

{{{(x)^2-(8)^2=(x+8)(x-8)}}} Plug in {{{A=x}}} and {{{B=8}}}.

So this shows us that {{{x^2-64}}} factors to {{{(x+8)(x-8)}}}.

In other words {{{x^2-64=(x+8)(x-8)}}} for all values of x.

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