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

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

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Answer 1 · 2017-03-04T11:08:49

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

Explanation:

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

$x^{4} (λ - λ^{2} - 20 λ^{3})$ $x^4(lambda-lambda^2-20lambda^3)$ . The roots of

$λ - λ^{2} - 20 λ^{3} = 0$ $lambda-lambda^2-20lambda^3=0$ are

$λ = {0, - \frac{1}{4}, \frac{1}{5}}$ $lambda={0, -1/4,1/5}$ or

$λ - λ^{2} - 20 λ^{3} = 20 λ (λ + \frac{1}{4}) (λ - \frac{1}{5})$ $lambda-lambda^2-20lambda^3=20lambda(lambda+1/4)(lambda-1/5)$

then

$x^{3} y - x^{2} y^{2} - 20 x y^{3} = 20 x^{4} λ (λ + \frac{1}{4}) (λ - \frac{1}{5})$ $x^ { 3} y - x ^ { 2} y ^ { 2} - 20x y ^ { 3} =20x^4lambda(lambda+1/4)(lambda-1/5)$ or

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

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

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

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Explanation:

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