How do you factor #x^3 -x^2 + x -6 = 0#?

1 Answer
Jun 9, 2016

Answer:

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

Derived by substitution and observation.

Explanation:

Given:#" "x^3-x^2+x-6#

Factors of 6 are {1,6} ; {2,3}
So #x# equals at least one of these factors is a solution.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Try #x=1#
#1-1+1-6!=0 color(red)(" Fail")#

Try #x=2#
#2^3-2^2+2-6=0 =>color(green)(x-2 " is a factor"#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So we have #(x-2)(?)=x^3-x^2+x-6=0#

Building by 1 term at a time.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("1st term:")# this needs to build #x^3#

#(x-2)(x^2+?) # gives #x^3#

But #-2xx x^2 = -2x^2# giving

#(x-2)(x^2+?) = x^3-2x^2+?#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("2nd term")# this build to be #-x^2#

To 'force' #-2x^2# into #-x^2# we need to generate #+x^2#

So our next number is #x# giving

#(x-2)(x^2+x+?) = x^3-2x^2+x^2-2x+?#

#(x-2)(x^2+x+?) = x^3-x^2-2x+?#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("3rd term")# The #x# from #x^2+x# generated is #-2x# but we need it to be #+x#. So to force this change we need to generate #+3x#

So our next number is 3 giving

#(x-2)(x^2+x+3+?) =x^3-x^2-2x+3x-6# and this is our starting equation so the two bracket factors are:

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

We have now built all the terms

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The latter bracket can not be split into any further whole number factorisations so we stop.

The only whole number factors of 3 is {1,3} and #3x-x!=+x#