How do you factor completely $3x^3y^2 - 3x^2y^2 + 3xy^2$ ?

Question

3244 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-07T19:37:27

$3x^3y^2-3x^2y^2+3xy^2$

$= 3xy^2(x^2-x+1)$

$= 3xy^2(x-1/2-sqrt(3)/2i)(x-1/2+sqrt(3)/2i)$

All of the terms are divisible by $3xy^2$ , so separate that out as a factor first:

$3x^3y^2-3x^2y^2+3xy^2 = 3xy^2(x^2-x+1)$

The remaining quadratic factor is of the form $ax^2+bx+c$ with $a=1$ , $b=-1$ and $c=1$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (-1)^2 - (4*1*1) = 1-4 = -3$

Since this is negative the quadratic has no factors with Real coefficients.

If we allow Complex coefficients then we can factor it further.

$x^2-x+1 = 0$ when

$x = (-b+-sqrt(b^2-4ac))/(2a)$
$=(-b+-sqrt(Delta))/(2a)$

$=(1+-sqrt(-3))/2$

$=1/2+-sqrt(3)/2i$

Hence we find:

$x^2-x+1 = (x-1/2-sqrt(3)/2i)(x-1/2+sqrt(3)/2i)$

How do you factor completely 3x^3y^2 - 3x^2y^2 + 3xy^23x^3y^2 - 3x^2y^2 + 3xy^2?