How do you factor completely: $42x^3y^4 - 63x^2y$ ?

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Answer 1 · 2015-09-03T16:31:04

$42x^3y^4-63x^2y = 21x^2y(2xy^3-3)$

The highest common factor of the terms is $21x^2y$ , so separate that out as a factor to get:

$42x^3y^4-63x^2y = 21x^2y(2xy^3-3)$

The remaining factor $(2xy^3-3)$ does not factorise further.

How do you factor completely: 42x^3y^4 - 63x^2y42x^3y^4 - 63x^2y?