Is it possible to factor #y= 6x^3-9x+3 #? If so, what are the factors?

1 Answer
George C.
Jan 9, 2016

Yes:

#y = 6x^3-9x+3#

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

#= 6(x-1)(x-1/2-sqrt(3)/2)(x-1/2+sqrt(3)/2)#

Explanation:

The difference of squares identity can be written:

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

We use this later.

First separate out the common scalar factor #3# to find:

#y = 6x^3-9x+3 = 3(2x^3-3x+1)#

Next note that the sum of the coefficients is zero. That is #2-3+1 = 0#. So #x=1# is a zero and #(x-1)# is a factor:

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

We can factor the remaining quadratic expression by completing the square and using the difference of squares identity...

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

#=2(x^2-x-1/2)#

#=2(x^2-x+1/4-3/4)#

#=2((x-1/2)^2-(sqrt(3)/2)^2)#

#=2((x-1/2)-sqrt(3)/2)((x-1/2)+sqrt(3)/2)#

#=2(x-1/2-sqrt(3)/2)(x-1/2+sqrt(3)/2)#

Putting it all together:

#y = 6x^3-9x+3#

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

#= 6(x-1)(x-1/2-sqrt(3)/2)(x-1/2+sqrt(3)/2)#

Related questions
See all questions in Monomial Factors of Polynomials
Impact of this question
1961 views around the world
You can reuse this answer
Creative Commons License