How do you factor completely $- 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3}$ $−12x^3y^5 − 9x^2y^2 + 12xy^3$ ?

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How do you factor completely $- 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3}$ $−12x^3y^5 − 9x^2y^2 + 12xy^3$ ?

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Answer 1 · 2017-04-13T07:22:40

$- 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3} = - 3 x y^{2} (4 x^{2} y^{3} + 3 x - 4 y)$ $-12x^3y^5-9x^2y^2+12xy^3 = -3xy^2(4x^2y^3+3x-4y)$

But note that:

$9 x^{4} y^{4} - 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3} = 3 x y^{2} (x y - 1) (x y + 1) (3 x - 4 y)$ $color(red)(9x^4y^4)-12x^3y^5-9x^2y^2+12xy^3 = 3xy^2(xy-1)(xy+1)(3x-4y)$

Explanation:

Given:

$- 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3}$ $-12x^3y^5-9x^2y^2+12xy^3$

First note that all of the terms are divisible by $3$ $3$ , $x$ $x$ and $y^{2}$ $y^2$ , and hence by $3 x y^{2}$ $3xy^2$ . So we can separate that out as a factor. I will use $- 3 x y^{2}$ $-3xy^2$ because I prefer positive leading coefficients:

$- 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3} = - 3 x y^{2} (4 x^{2} y^{3} + 3 x - 4 y)$ $-12x^3y^5-9x^2y^2+12xy^3 = -3xy^2(4x^2y^3+3x-4y)$

It is not possible to factor this further.

I suspect that a leading term $9 x^{4} y^{4}$ $9x^4y^4$ is missing from the question, since we find:

$9 x^{4} y^{4} - 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3} = 3 x y^{2} (3 x^{3} y^{2} - 4 x^{2} y^{3} - 3 x + 4 y)$ $9x^4y^4-12x^3y^5-9x^2y^2+12xy^3 = 3xy^2(3x^3y^2-4x^2y^3-3x+4y)$

$9 x^{4} y^{4} - 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3} = 3 x y^{2} ((3 x^{3} y^{2} - 4 x^{2} y^{3}) - (3 x - 4 y))$ $color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2((3x^3y^2-4x^2y^3)-(3x-4y))$

$9 x^{4} y^{4} - 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3} = 3 x y^{2} (x^{2} y^{2} (3 x - 4 y) - 1 (3 x - 4 y))$ $color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2(x^2y^2(3x-4y)-1(3x-4y))$

$9 x^{4} y^{4} - 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3} = 3 x y^{2} (x^{2} y^{2} - 1) (3 x - 4 y)$ $color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2(x^2y^2-1)(3x-4y)$

$9 x^{4} y^{4} - 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3} = 3 x y^{2} ({(x y)}^{2} - 1^{2}) (3 x - 4 y)$ $color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2((xy)^2-1^2)(3x-4y)$

$9 x^{4} y^{4} - 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3} = 3 x y^{2} (x y - 1) (x y + 1) (3 x - 4 y)$ $color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2(xy-1)(xy+1)(3x-4y)$

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How do you factor completely $- 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3}$ $−12x^3y^5 − 9x^2y^2 + 12xy^3$ ?

1 Answer

Explanation:

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How do you factor completely −12x^3y^5 − 9x^2y^2 + 12xy^3−12x3y5−9x2y2+12xy3−12x^3y^5 − 9x^2y^2 + 12xy^3?

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How do you factor completely $- 12 x^{3} y^{5} - 9 x^{2} y^{2} + 12 x y^{3}$ $−12x^3y^5 − 9x^2y^2 + 12xy^3$ ?