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)#