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

1 Answer
Oct 1, 2015

Answer:

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

Explanation:

Given
#color(white)("XXX")6x^3-6x^2+18x#

Extract the obvious common factor #6x# from each term:
#color(white)("XXX")6x(x^2-x+3)#

We could try to factor #(x^2-x+3)#
but an examination of the discriminant (see below) gives us
#color(white)("XXX")#Delta = (-1)^2-4(1)(3) < 0#
which tells us there are no Real roots;
so factors are not available.

Discriminants and their significance
#color(white)("XXX")#for a quadratic in the form
#color(white)("XXXXX")ax^2+bx+c#
#color(white)("XXX")#the discriminant is
#color(white)("XXXXX")Delta = b^2-4ac#

#color(white)("XXX")Delta { (< 0 rarr "no Real roots"), (= 0 rarr "exactly one Real root"), (> 0 rarr "two Real roots"):}#