Question 263196


{{{4w^2-225}}} Start with the given expression.



{{{(2w)^2-225}}} Rewrite {{{4w^2}}} as {{{(2w)^2}}}.



{{{(2w)^2-(15)^2}}} Rewrite {{{225}}} as {{{(15)^2}}}.



Notice how we have a difference of squares {{{A^2-B^2}}} where in this case {{{A=2w}}} and {{{B=15}}}.



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.



{{{(2w)^2-(15)^2=(2w+15)(2w-15)}}} Plug in {{{A=2w}}} and {{{B=15}}}.



So this shows us that {{{4w^2-225}}} factors to {{{(2w+15)(2w-15)}}}.



In other words {{{4w^2-225=(2w+15)(2w-15)}}}.