Question

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

Answer 1

`(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`