Question 87365
*[invoke change_this_name4450 1, 8, 15]


{{{12x^3-3xy^2}}}


{{{3x(4x^2-y^2)}}} Factor out the GCF


Now lets factor {{{4x^2-y^2}}} using the difference of squares


{{{4x^2-1y^2}}} Start with the given expression


Let {{{A^2=4x^2}}} and {{{B^2=1y^2}}}. So we get this:


{{{4x^2-1y^2=A^2-B^2}}}


Since {{{A^2=4x^2}}}, A can be solved for:

{{{sqrt(A^2)=sqrt(4x^2)}}} Take the square root of both sides


{{{A=2x}}}


Since {{{B^2=1y^2}}}, B can be solved for:

{{{sqrt(B^2)=sqrt(1y^2)}}} Take the square root of both sides


{{{B=1y}}}


Since we have a difference of squares, we can factor it like this:


{{{A^2-B^2=(A+B)(A-B)}}}




Now replace A and B

{{{4x^2-1y^2=(2x+1y)(2x-1y)}}} Plug in {{{A=2x}}} and {{{B=1y}}}


So the expression


{{{4x^2-1y^2}}}


factors to


{{{(2x+1y)(2x-1y)}}}


Notice that if you foil the factored expression, you get the original expression. This verifies our answer.


So {{{12x^3-3xy^2}}} factors to 



{{{3x(2x+y)(2x-y)}}}