<s>[INST] Factoring polynomials of degree 8 A question in my book asks me to factor $x^8+3x^3-1$ and in the answers, it gives me the correct factorization but gives no explanation on how this was done. Can someone explain how this is factored, Thanks.
 [/INST] Easy enough to try all possible factors into two parts. We know the first two fail, there are no rational roots.
In each of seven cases, we make conclusions both from the left and from the right, as we will be getting the $x^7$ and $x^6$ terms to have coefficient zero on the left, also the $x^2$ and $x$ terms. We do not write out the whole messy system at once. 
For example
$ \left( x^2 + ax - 1 \right)  \left( x^6 + b x^5 + c x^4 + d x^3 + e x^2 + fx +1 \right)   $ requires $a+b=0,$ then $c +ab -1 = 0.$ So far $b =-a, c = 1 + a^2.$ On the right end, $a-f=0,$ then $1 + af -e = 0,$ so $f=a, e = 1 + a^2.$ Continue with
$$ \left( x^2 + ax - 1 \right)  \left( x^6 -a x^5 + (1+a^2) x^4 + d x^3 + (1+a^2) x^2 + ax +1 \right)   $$
and see whether $a$ and $d$ come out as integers.
Indeed, after correcting $c,$ I get
$$ x^8 + (2a+a^3 + d)x^5 + ad x^4 + (2a+a^3 - d)x^3 - 1.  $$
This means that $ad = 0.$ From $2a+a^3 + d =0,$ this means that both $a,d = 0$ or $a$ is irrational, which means this example fails with $2a+a^3 - d=0 \neq 3.$
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$$ \left( x-1 \right)  \left( x^7 + a x^6 + b x^5 + c x^4 + d x^3 + e x^2 + fx +1 \right)   $$
$$ \left( x+1 \right)  \left( x^7 + a x^6 + b x^5 + c x^4 + d x^3 + e x^2 + fx -1 \right)   $$
$$ \left( x^2 + ax - 1 \right)  \left( x^6 + b x^5 + c x^4 + d x^3 + e x^2 + fx +1 \right)   $$
$$ \left( x^2 + ax + 1 \right)  \left( x^6 + b x^5 + c x^4 + d x^3 + e x^2 + fx - 1 \right)   $$
$$ \left( x^3 + a x^2 + bx -1 \right)  \left( x^5 + c x^4 + d x^3 + e x^2 + fx +1 \right)   $$
$$ \left( x^3 + a x^2 + bx +1 \right)  \left( x^5 + c x^4 + d x^3 + e x^2 + fx -1 \right)   $$
$$ \left( x^4 + a x^3 + b x^2 + c x - 1 \right)  \left( x^4 + d x^3 + e x^2 + fx + 1 \right)   $$
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