How do you factor #x^ { 3} y - x ^ { 2} y ^ { 2} - 20x y ^ { 3}# completely?

Question

Eva Abramson · Accepted Answer

#x^ { 3} y - x ^ { 2} y ^ { 2} - 20x y ^ { 3} =xy(4y+x)(5y-x)#

This is a homogeneous formula so making #y = lambda x# and substituting we have

#x^4(lambda-lambda^2-20lambda^3)#. The roots of

#lambda-lambda^2-20lambda^3=0# are

#lambda={0, -1/4,1/5}# or

#lambda-lambda^2-20lambda^3=20lambda(lambda+1/4)(lambda-1/5)#

then

#x^ { 3} y - x ^ { 2} y ^ { 2} - 20x y ^ { 3} =20x^4lambda(lambda+1/4)(lambda-1/5)# or

#x^ { 3} y - x ^ { 2} y ^ { 2} - 20x y ^ { 3} =20xy(y+x/4)(y-x/5)# or

#x^ { 3} y - x ^ { 2} y ^ { 2} - 20x y ^ { 3} =xy(4y+x)(5y-x)#

Nathan Axel · Answer

undefined To factor the expression $ x^3y - x^2y^2 - 20xy^3 $ completely, you can first factor out the common factor $ xy $, then factor the remaining expression as a quadratic trinomial in terms of $ x $. The factored expression is $ xy(x - 5y)(x + y) $.