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

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Answer 1 · 2017-03-04T11:08:49

#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)#