Question 207959
{{{n^4-4n^3+n-4}}} Start with the given expression



{{{(n^4-4n^3)+(n-4)}}} Group like terms



{{{n^3(n-4)+1(n-4)}}} Factor out the GCF {{{n^3}}} out of the first group. Factor out the GCF {{{1}}} out of the second group



{{{(n^3+1)(n-4)}}} Since we have the common term {{{n-4}}}, we can combine like terms



{{{(n+1)(n^2-n+1)(n-4)}}} Now factor {{{n^3+1}}} to get {{{(n+1)(n^2-n+1)}}} (hint: use the sum of cubes formula)



{{{(n+1)(n-4)(n^2-n+1)}}} Rearrange the terms.



So {{{n^4-4n^3+n-4}}} completely factors to {{{(n+1)(n-4)(n^2-n+1)}}}



In other words, {{{n^4-4n^3+n-4=(n+1)(n-4)(n^2-n+1)}}}