How do you factor $x^{5} + x^{3} + x^{2} + 1$ $x^5+x^3+x^2+1$ ?

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How do you factor $x^{5} + x^{3} + x^{2} + 1$ $x^5+x^3+x^2+1$ ?

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Answer 1 · 2015-04-17T11:39:29

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

Using synthetic division
$(x^{5} + x^{3} + x^{2} + 1) \div (x + 1)$ $(x^5+x^3+x^2+1)div (x+1)$
yields the factors
$(x + 1) (x^{4} - x^{3} + x - 1)$ $(x+1)(x^4-x^3+x-1)$

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

Using synthetic division again
$(x^{4} - x^{3} + x - 1) \div (x - 1)$ $(x^4-x^3+x-1)div(x-1)$
gives the factors
$x^{4} - x^{3} + x - 1) = (x - 1) (x^{3} + 1)$ $x^4-x^3+x-1) = (x-1)(x^3+1)$

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

The complete factorization is
$x^{5} + x^{3} + x^{2} + 1$ $x^5+x^3+x^2+1$

$= {(x + 1)}^{2} (x - 1) (x^{2} - x + 1)$ $= (x+1)^2(x-1)(x^2-x+1)$

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