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So I found this question: Factor $x^{14}+8x^{13}+3$ over the rationals And I was reading the solution, but I am not understanding some of the parts and would really appreciate further explanation.

Specifically, when we reduce our polynomial mod 3, to get: $f(x)g(x)=x^{13}(x+2)$

We then say that $f(x)=x^r$ and $g(x)=x^{13−r}(x+2)$ and "one of the constant terms of f, g is not a multiple of 3, and so r=0 or 13"

Can someone explain what the part in bold means? so g is a constant term and not a multiple of 3? how do we conclude that r = 0 or 13?

Also, how did we assert that our polynomial has no linear factor g in $ℤ[x]$? Which part of the argument does this follow from?

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    The "constant term" of a polynomial is the term of degree zero. For example, the constant term of $x^2 + 5x +4$ is $4$. The constant term of $x^3$ is $0$. Here you know that the constant terms of $f$ and $g$ cannot both be multiples of $3$,, because otherwise, the constant term of $fg$, which is the product of the constant terms of $f$ and $g$, would have to be a multiple of $9$, which is not the case. – 2017-02-22
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    Now if $r>0$, then the constant term of $f$ must be a multiple of $3$. Likewise, if $13 - r > 0$, then the constant term of $g$ must be a multiple of $3$. – 2017-02-22
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    Regarding the question why the polynomial has no linear factors, this follows from Gauss Lemma: Let $\frac{x}{y} \in \mathbb Q$ be a rational root of polynomial $\sum_{i=0}^{n}{a_iX^i}$ with coefficients in $\mathbb Z$. Then $x | a_0$ and $y | a_n$. So in your case a root can only be $\pm 1, \pm 3$. Just plug all of them in and see that they are no roots. – 2017-02-22

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I shall use the notations from the linked post. Notice that, if $f(x) g(x) = x^{14} + 8x^{13} + 3$, it follows that $f(0) g(0) = 3$ whence, since $f,g \in \Bbb Z[x]$, it follows that either $f(0) = \pm 1$ and $g(0) = \pm 3$, or that $f(0) = \pm 3$ and $g(0) = \pm 1$.

In the first case, we have $g(0) \equiv 0 \pmod 3$. Can this happen if $r=12$? No, because in this case you would have $g(0) \equiv 2 \pmod 3$. It remains that $r=0$.

In the second case, we have $f(-) \equiv 0 \pmod 3$. Can this happen if $r=0$? No, because in this case you would have $f(0) \equiv 1 \pmod 3$. It remains that $r=13$.

We conclude that either $r=0$ or $r=13$.

To answer your second question, imagine that $x^{14} + 8x^{13} + 3$ had a linear factor $ax-b$ in $\Bbb Z[x]$. Then there exist a polynomial $cx^{13} + \dots \in \Bbb Z[x]$ such that $x^{14} + 8x^{13} + 3 = (ax-b) (cx^{13} + \dots)$, which implies that $ax \cdot cx^{13} = x^{14}$, i.e. that $ac = 1$. Since $a,c \in \Bbb Z$, it follows that $a = \pm 1$.

Let us suppose that $a=1$. It follows that $x-b \mid x^{14} + 8x^{13} + 3$, i.e. that $b$ is a root of $x^{14} + 8x^{13} + 3$, i.e. that $b^{14} + 8b^{13} + 3 = 0$, i.e. that $b^{13}(b+8) = -3$, whence it follows that $b^{13} \mid -3$, and the only integer solution of this is $b = \pm 1$. Check for yourself, though, that $+1$ and $-1$ are not roots of $x^{14} + 8x^{13} + 3$.

A similar analysis can be carried over if $a=-1$.

We conclude that $x^{14} + 8x^{13} + 3$ has no linear factor in $\Bbb Z[x]$.