How do you factor completely: 10x^4y^3 − 5x^3y^2 + 20x^2y10x4y35x3y2+20x2y?

1 Answer
Jul 23, 2015

10x^4y^3-5x^3y^2+20x^2y10x4y35x3y2+20x2y

=5x^2y(2(xy)^2-(xy)+4)=5x2y(2(xy)2(xy)+4)

=10x^2y(xy - 1/4 -i sqrt(31)/4)(xy - 1/4 + i sqrt(31)/4)=10x2y(xy14i314)(xy14+i314)

Explanation:

10x^4y^3-5x^3y^2+20x^2y10x4y35x3y2+20x2y

=5x^2y(2x^2y^2-xy+4)=5x2y(2x2y2xy+4)

=5x^2y(2(xy)^2-(xy)+4)=5x2y(2(xy)2(xy)+4)

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

The roots of 2t^2-t+4 = 02t2t+4=0 are given by the quadratic formula as:

t = (1+-sqrt(1^2-(4xx2xx4)))/(2*2)t=1±12(4×2×4)22

=(1+-sqrt(-31))/4=1±314

=1/4+-i sqrt(31)/4=14±i314

So if we allow complex coefficients we get:

2(xy)^2-(xy)+42(xy)2(xy)+4

= 2(xy - 1/4 -i sqrt(31)/4)(xy - 1/4 + i sqrt(31)/4)=2(xy14i314)(xy14+i314)

Hence:

10x^4y^3-5x^3y^2+20x^2y10x4y35x3y2+20x2y

=10x^2y(xy - 1/4 -i sqrt(31)/4)(xy - 1/4 + i sqrt(31)/4)=10x2y(xy14i314)(xy14+i314)