How do you factor #a^5 - 3a^4 + a^3#?

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Answer 1 · 2015-06-22T17:10:59

#a^5-3a^4+a^3#

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

#=a^3(a-(3+sqrt(5))/2)(a-(3-sqrt(5))/2)#

First separate out the common factor #a^3# to get:

#a^5-3a^4+a^3=a^3(a^2-3a+1)#

To find the factors of #f(a)=a^2-3a+1#, use the quadratic formula to find the roots of #f(a) = 0#, as:

#a = (3+-sqrt(3^2-(4xx1xx1)))/(2xx1)#

#=(3+-sqrt(9-4))/2#

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

Hence #f(a)# can be factored as:

#f(a) = (a-(3+sqrt(5))/2)(a-(3-sqrt(5))/2)#