How do you factor #n^ { 3} - 8n ^ { 2} + 16n = 0#?

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Answer 1 · 2017-10-19T12:34:40

#n(n-4)^2=0#

start by taking out any common factors

#n^3-8n^2+16n=0#

=>#n(n^2-8n+16)=0#

we have a quadratic expression in the brackets which factorises

we need factors of #+16# that sum to #-8#

they will be

#-4,-4#

so we end up with

#n(n-4)(n-4)=0#

#n(n-4)^2=0#

if you then what to solve

we have #n=0, n=4#