Question 190750


{{{81x^2-121}}} Start with the given expression.



{{{(9x)^2-121}}} Rewrite {{{81x^2}}} as {{{(9x)^2}}}.



{{{(9x)^2-(11)^2}}} Rewrite {{{121}}} as {{{(11)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=9x}}} and {{{B=11}}}.



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.



{{{(9x)^2-(11)^2=(9x+11)(9x-11)}}} Plug in {{{A=9x}}} and {{{B=11}}}.



So this shows us that {{{81x^2-121}}} factors to {{{(9x+11)(9x-11)}}}.



In other words {{{81x^2-121=(9x+11)(9x-11)}}}.