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

1 Answer
Feb 16, 2016

Answer:

Factors are hence #(x-2)(x^2+3)#

Explanation:

To factorize the expression #x^3−2x^2+3x−6#, we first identify a value of #x# for which the value of function is #0#. This could be a factor of last term i.e. #-6# i.e. among #(1, 2, 3, 6, -1, -2, -3, -6)#.

It is seen that for #x-2# function is zero. Hence #(x-2)# is a factor of #x^3−2x^2+3x−6#. Dividing latter by former, we get the factors of the function as

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

Now as the determinant (#b^2-4ac# if the function is #ax^2+bx+c#) of #x^2+3# is

#0^2-4.1.3# = #-12#. a negative number, this cannot be factorized into rational factors (assuming that to be a condition).

The factors are hence #(x-2)(x^2+3)#