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

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

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