How do you factor #x^5+x^3+x^2+1#?

1 Answer
Alan P.
Apr 17, 2015

Since #x^n# is negative for odd values of #n#
an obvious "solution" (if we pretend this expression #=0#)
is #x=-1#
Therefore one of the factors is #(x+1)#

Using synthetic division
#(x^5+x^3+x^2+1)div (x+1)#
yields the factors
#(x+1)(x^4-x^3+x-1)#

Again for #(x^4-x^3+x-1) = 0#
we have an obvious solution, #x=1#
So #(x-1)# is a factor

Using synthetic division again
#(x^4-x^3+x-1)div(x-1)#
gives the factors
#x^4-x^3+x-1) = (x-1)(x^3+1)#

Once again
#(x^3+1)=0#
has an obvious solution
#x=-1#
which leads to factors
#(x^3-1) = (x+1)(x^2-x+1)#

The complete factorization is
#x^5+x^3+x^2+1#

#= (x+1)^2(x-1)(x^2-x+1)#

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