How do you factor #6x^3-6x^2-x+1#?

2 Answers
hev1
Mar 23, 2018

#6x^3 - 6x^2 - x + 1#
#=x(6x^2 - 6x -1 + 1)#
#=x(6x^2 - 6x)#
#=6x(x^2 - x)#

George C.
Mar 23, 2018

#6x^3-6x^2-x+1 = (6x^2-1)(x-1)#

#color(white)(6x^3-6x^2-x+1) = (sqrt(6)x-1)(sqrt(6)x+1)(x-1)#

Explanation:

This cubic quadrinomial factors by grouping and using the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with #A=sqrt(6)x# and #B=1# as follows:

#6x^3-6x^2-x+1 = (6x^3-6x^2)-(x-1)#

#color(white)(6x^3-6x^2-x+1) = 6x^2(x-1)-1(x-1)#

#color(white)(6x^3-6x^2-x+1) = (6x^2-1)(x-1)#

#color(white)(6x^3-6x^2-x+1) = ((sqrt(6)x)^2-1^2)(x-1)#

#color(white)(6x^3-6x^2-x+1) = (sqrt(6)x-1)(sqrt(6)x+1)(x-1)#

Related questions
See all questions in Factoring by Grouping
Impact of this question
2335 views around the world
You can reuse this answer
Creative Commons License