Question 156111
{{{5u^3-80uy^2}}} Start with the given expression



{{{5u(u^2-16y^2)}}} Factor out the GCF {{{5u}}}




{{{5u((u)^2-16y^2)}}} Rewrite {{{u^2}}} as {{{(u)^2}}}.



{{{5u((u)^2-(4y)^2)}}} Rewrite {{{16y^2}}} as {{{(4y)^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:



{{{5u(u+4y)(u-4y)}}} Factor {{{u^2-16y^2}}} to get {{{(u+4y)(u-4y)}}} by using the difference of squares.



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Answer:


So {{{5u^3-80uy^2}}} factors to {{{5u(u+4y)(u-4y)}}}