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

Question

Eva Abramson · Accepted Answer

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

Ethan Adkins · Answer

undefined To factor the expression $x^3 - 2x^2 + 3x - 6$, you can first look for any common factors among the terms, then use techniques such as grouping or synthetic division. In this case, the expression can be factored as follows:

$x^3 - 2x^2 + 3x - 6 = (x - 2)(x^2 + 3)$