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

Question

Savannah Anderson · Accepted Answer

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

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

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)#

Isaiah Anthony · Answer

undefined To factor completely the expression $ -6m^5 + 34m^3 - 40m $, first, we can factor out the greatest common factor, which is 2m:

$$ 2m(-3m^4 + 17m^2 - 20) $$

Now, we focus on factoring the quadratic expression inside the parentheses, $ -3m^4 + 17m^2 - 20 $. This can be factored as:

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

Further, we can factor $ 3m^2 - 5 $ as $ (3m^2 - 5) = (3m - \sqrt{5})(3m + \sqrt{5}) $, and $ m^2 - 4 $ as $ (m^2 - 4) = (m - 2)(m + 2) $.

Therefore, the expression $ -6m^5 + 34m^3 - 40m $ can be factored completely as:

$$ 2m(3m - \sqrt{5})(3m + \sqrt{5})(m - 2)(m + 2) $$