How do you factor completely #-6m^5+34m^3-40m#?

1 Answer
George C.
Jul 2, 2016

#color(white)()#
#-6m^5+34m^3-40m#

#= -2m(sqrt(3)m-sqrt(5))(sqrt(3)m+sqrt(5))(m-2)(m+2)#

Explanation:

First note that all of the terms are divisible by #-2m#. So we can separate that out as a factor:

#-6m^5+34m^3-40m = -2m(3m^4-17m^2+20)#

The remaining quartic expression can be treated as a quadratic in #m^2#.

We can use an AC method to find quadratic factors:

Find a pair of factors of #AC = 3*20 = 60# with sum #B=17#.

The pair #12, 5# works, so we can use that to split the middle term and factor by grouping:

#3m^4-17m^2+20 = (3m^4-12m^2)-(5m^2-20)#

#=3m^2(m^2-4)-5(m^2-4)#

#=(3m^2-5)(m^2-4)#

#=(3m^2-5)(m-2)(m+2)#

#=(sqrt(3)m-sqrt(5))(sqrt(3)m+sqrt(5))(m-2)(m+2)#

Putting it all together:

#-6m^5+34m^3-40m#

#= -2m(sqrt(3)m-sqrt(5))(sqrt(3)m+sqrt(5))(m-2)(m+2)#

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