How do you factor #21x^3 - 18x^2y + 24xy^2#?

1 Answer
Sep 19, 2016

Answer:

#21x^3-18x^2y+24xy^2 = 3x(7x^2-6xy+8y^2)#

Explanation:

Note that all of the terms are divisible by #3x#, so we can separate that out as a factor first:

#21x^3-18x^2y+24xy^2 = 3x(7x^2-6xy+8y^2)#

The remaining quadratic is in the form #ax^2+bxy+cy^2# with #a=7#, #b=-6# and #c=8#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-6)^2-4*7*8 = 36 - 224 = -188#

Since #Delta < 0# this quadratic has no linear factors with Real coefficients.

#color(white)()#
Complex coefficients

If we are determined to factor it, we can find some Complex coefficients using the quadratic formula to find zeros:

#x/y = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x/y) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x/y) = (6+-sqrt(-188))/14#

#color(white)(x/y) = (6+-2sqrt(47)i)/14#

#color(white)(x/y) = (3+-sqrt(47)i)/7#

Hence:

#7x^2-6xy+8y^2 = 7(x - ((3+sqrt(47)i)/7)y)(x - ((3-sqrt(47)i)/7)y)#

#color(white)(7x^2-6xy+8y^2) = 1/7(7x - (3+sqrt(47)i)y)(7x - (3-sqrt(47)i)y)#