How do you factor # 4x^3 - x^2 -12x + 3# by grouping?

1 Answer
Jun 23, 2016

#(4x-1)(x^2-3)#

Explanation:

Think of this cubic as two groups:

#overbrace((4x^3-x^2))^"Group 1"+overbrace((-12x+3))^"Group 2"#

We will want to find a common factor in each group. From #4x^3-x^2#, we see a common factor of #x^2#:

#overbrace(x^2(4x-1))^"Group 1"+overbrace((-12x+3))^"Group 2"#

From Group 2, we could either factor out a #+3# or a #-3#. To make the leading term positive, we will factor out a #-3# from both #-12x# and #3#, leaving:

#overbrace(x^2(4x-1))^"Group 1"+overbrace(-3(4x-1))^"Group 2"#

From here, notice that there is a common factor between Group 1 and Group 2. Both have terms, #x^2# and #-3#, which are being multiplied by the other term #(4x-1)#. Here, #(4x-1)#, despite it being made up of two terms, is the common factor and can be factored out. The #x^2# and #-3# are combined since they share this common factor:

#(4x-1)(x^2-3)#

Depending on your level of instruction, you may recognize that #x^2-3# is a difference of squares, and can be factorized yet. If not, this is a fine final answer.