How do you factor completely $2x^3y^4 − 8x^2y^3 + 6xy^2$ ?

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Answer 1 · 2016-08-18T19:41:31

$2x^3y^4-8x^2y^3+6xy^2=2xy^2(xy-3)(xy-1)$

Separate out the common factor $2xy^2$ of the terms, then factor as a quadratic in $xy$ as follows:

$2x^3y^4-8x^2y^3+6xy^2$

$=2xy^2(x^2y^2-4xy+3)$

$=2xy^2((xy)^2-4(xy)+3)$

$=2xy^2(xy-3)(xy-1)$

How do you factor completely 2x^3y^4 − 8x^2y^3 + 6xy^22x^3y^4 − 8x^2y^3 + 6xy^2?