How do you factor # x^3-3x^2+6x-18#?

1 Answer
Sep 3, 2016

Answer:

#x^3-3x^2+6x-18 = (x^2+6)(x-3)#

#color(white)(x^3-3x^2+6x-18) = (x-sqrt(6)i)(x+sqrt(6)i)(x-3)#

Explanation:

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms, so this will factor by grouping:

#x^3-3x^2+6x-18 = (x^3-3x^2)+(6x-18)#

#color(white)(x^3-3x^2+6x-18) = x^2(x-3)+6(x-3)#

#color(white)(x^3-3x^2+6x-18) = (x^2+6)(x-3)#

That's as far as you can go using Real coefficients since #x^2+6 >= 6 > 0# for any Real value of #x#. If you allow Complex coefficients then this factors further as:

#color(white)(x^3-3x^2+6x-18) = (x^2-(sqrt(6)i)^2)(x-3)#

#color(white)(x^3-3x^2+6x-18) = (x-sqrt(6)i)(x+sqrt(6)i)(x-3)#