How do you factor $6x^4y^3 + 21x^3y^2 − 9x^2y$ ?

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Answer 1 · 2016-04-24T18:58:13

$6x^4y^3+21x^3y^2-9x^2y$

$=3x^2y(2x^2y^2+7xy-3)$

$=6x^2y(xy+7/4-sqrt(73)/4)(xy+7/4+sqrt(73)/4)$

First notice that all of the terms are divisible by $3x^2y$ , so separate that out as a factor:

$6x^4y^3+21x^3y^2-9x^2y=3x^2y(2x^2y^2+7xy-3)$

The remaining factor is a quadratic in $xy$ , whose zeros we can find using the quadratic formula:

$xy = (-7+-sqrt(7^2-(4*2*-3)))/(2*2)$

$=(-7+-sqrt(49+24))/4$

$=(-7+-sqrt(73))/4$

Hence:

$2x^2y^2+7xy-3 = 2(xy+7/4-sqrt(73)/4)(xy+7/4+sqrt(73)/4)$

Putting it all together:

$6x^4y^3+21x^3y^2-9x^2y$

$=3x^2y(2x^2y^2+7xy-3)$

$=6x^2y(xy+7/4-sqrt(73)/4)(xy+7/4+sqrt(73)/4)$

How do you factor 6x^4y^3 + 21x^3y^2 − 9x^2y6x^4y^3 + 21x^3y^2 − 9x^2y?