How do you factor #4c^3-2c^2-6c#?

1 Answer
George C.
Aug 30, 2016

#4c^3-2c^2-6c=c(2c-3)(2c+2)#

Explanation:

Note that all of the terms are divisible by #c#, so we can separate that out as a factor first.

We can then factor the remaining quadratic by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(2c-1/2)# and #b=5/2# as follows:

#4c^3-2c^2-6c#

#=c(4c^2-2c-6)#

#=c((2c-1/2)^2-1/4-6)#

#=c((2c-1/2)^2-25/4)#

#=c((2c-1/2)^2-(5/2)^2)#

#=c((2c-1/2)-5/2)((2c-1/2)+5/2)#

#=c(2c-3)(2c+2)#

Related questions
See all questions in Factoring Completely
Impact of this question
1241 views around the world
You can reuse this answer
Creative Commons License