How do you factor $3 x^{4} y - 24 x y^{4}$ $3x^4 y - 24xy^4$ ?

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How do you factor $3 x^{4} y - 24 x y^{4}$ $3x^4 y - 24xy^4$ ?

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Answer 1 · 2015-04-10T18:43:37

First remove the common factot (s). Then see whether you're finished.

$3 x^{4} y - 24 x y^{4} = 3 x y (x^{3} - 8 y^{3})$ $3x^4y-24xy^4 = 3xy (x^3-8 y^3)$

We're not finished because 8 is a cube, so $8 y^{3} = {(2 y)}^{3}$ $8y^3= (2y)^3$ and we can factor the difference of cubes: $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3 =(a-b)(a^2+ab+b^2)$

So we get
$3 x^{4} y - 24 x y^{4} = 3 x y (x^{3} - 8 y^{3}) = 3 x y (x - 2 y) (x^{2} + 2 x y + 4 y^{2})$ $3x^4y-24xy^4=3xy (x^3-8y^3)= 3xy (x-2y)(x^2+2xy+4y^2)$

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How do you factor $3 x^{4} y - 24 x y^{4}$ $3x^4 y - 24xy^4$ ?

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