Is it possible to factor #y=x^3 - 3x^2 - 4x + 12 #? If so, what are the factors?

1 Answer
Jan 20, 2016

Answer:

(x - 3 )(x - 2 )(x + 2 )

Explanation:

By factorising in pairs :

#[ x^2 ( x - 3 ) ] + [ - 4 ( x - 3 ) ]#

ie # x^2 (x - 3 ) - 4 ( x - 3 ) #

now there is a common factor of (x - 3 )

# (x - 3 )(x^2 - 4 ) #

# x^2 - 4 # is a difference of 2 squares.

and # x^2 - 4 = ( x - 2 )(x + 2 ) #

#rArr x^3 -3x^2 - 4x + 12 = (x - 3 )(x - 2 )(x + 2 ) #