How do you factor completely: #10x^4y^3 − 5x^3y^2 + 20x^2y#?

1 Answer
Jul 23, 2015

Answer:

#10x^4y^3-5x^3y^2+20x^2y#

#=5x^2y(2(xy)^2-(xy)+4)#

#=10x^2y(xy - 1/4 -i sqrt(31)/4)(xy - 1/4 + i sqrt(31)/4)#

Explanation:

#10x^4y^3-5x^3y^2+20x^2y#

#=5x^2y(2x^2y^2-xy+4)#

#=5x^2y(2(xy)^2-(xy)+4)#

This is as far as we can go with real coefficients.

The roots of #2t^2-t+4 = 0# are given by the quadratic formula as:

#t = (1+-sqrt(1^2-(4xx2xx4)))/(2*2)#

#=(1+-sqrt(-31))/4#

#=1/4+-i sqrt(31)/4#

So if we allow complex coefficients we get:

#2(xy)^2-(xy)+4#

#= 2(xy - 1/4 -i sqrt(31)/4)(xy - 1/4 + i sqrt(31)/4)#

Hence:

#10x^4y^3-5x^3y^2+20x^2y#

#=10x^2y(xy - 1/4 -i sqrt(31)/4)(xy - 1/4 + i sqrt(31)/4)#