Question 168155


{{{9x^2-4y^8}}} Start with the given expression.



{{{(3x)^2-4y^8}}} Rewrite {{{9x^2}}} as {{{(3x)^2}}}.



{{{(3x)^2-(2y^4)^2}}} Rewrite {{{4y^8}}} as {{{(2y^4)^2}}}.



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



{{{(3x+2y^4)(3x-2y^4)}}} Factor the expression using the difference of squares.



So {{{9x^2-4y^8}}} factors to {{{(3x+2y^4)(3x-2y^4)}}}.



In other words {{{9x^2-4y^8=(3x+2y^4)(3x-2y^4)}}}.