Question 196925


{{{16x^2-81}}} Start with the given expression.



{{{(4x)^2-81}}} Rewrite {{{16x^2}}} as {{{(4x)^2}}}.



{{{(4x)^2-(9)^2}}} Rewrite {{{81}}} as {{{(9)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=4x}}} and {{{B=9}}}.



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.



{{{(4x)^2-(9)^2=(4x+9)(4x-9)}}} Plug in {{{A=4x}}} and {{{B=9}}}.



So this shows us that {{{16x^2-81}}} factors to {{{(4x+9)(4x-9)}}}.



In other words {{{16x^2-81=(4x+9)(4x-9)}}}.