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

##### 1 Answer

Jul 2, 2016

#= -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

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

The remaining quartic expression can be treated as a quadratic in

We can use an AC method to find quadratic factors:

Find a pair of factors of

The pair

#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)#