How do you factor completely: #6x^4-9x^3-36x^2+54x#?

2 Answers
George C.
Jul 12, 2015

Factor by grouping to get:

#6x^4-9x^3-36x^2+54x#

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

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

Explanation:

First, note that every term is divisible by #3x#, so

#6x^4-9x^3-36x^2+54x = 3x(2x^3-3x^2-12x+18)#

Next, to factor #2x^3-3x^2-12x+18#, notice that the ratio of the 1st and 2nd terms is the same as the ratio of the 3rd and 4th terms. So factoring by grouping will work:

#2x^3-3x^2-12x+18#

#=(2x^3-3x^2)-(12x-18)#

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

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

Next, use the difference of squares identity:

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

with #a=x# and #b=sqrt(6)# to get:

#(x^2-6)(2x-3)#

#=(x^2-sqrt(6)^2)(2x-3)#

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

Meave60
Jul 12, 2015

The answer is #3x(2x-3)(x+sqrt6)(x-sqrt6)#.

Explanation:

Factor by grouping.

#(6x^4-9x^3)-(36x^2-54x)#

Factor out the GCF for each group.

#3x^3(2x-3)-18x(2x-3)#

Factor out #(2x-3)#.

#(3x^3-18x)(2x-3)#

#(3x^3-18x)# can be factored further.

#3x(x^2-6)#

Substitute it back into the previous factorization.

#3x(x^2-6)(2x-3)#

#(x^2-6)# can be factored further.

#(x+sqrt6)(x-sqrt6)#

Put all of the factors together.

#3x(2x-3)(x+sqrt6)(x-sqrt6)#

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