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

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Answer 1 · 2015-07-23T16:14:37

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

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

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

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

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

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

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:

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

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

How do you factor completely: 10x^4y^3 − 5x^3y^2 + 20x^2y10x^4y^3 − 5x^3y^2 + 20x^2y?