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

##### 1 Answer

#### Explanation:

Note that all of the terms are divisible by

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

The remaining quadratic is in the form

This has discriminant

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

Since

**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)#