Factorizations versus integer solutions in single-variable quartics If $x$ is a positive integer, and I have the factorization
$$(x-1)(79x^3+159x^2-513x+255)=0,$$
what is the easiest way to conclude that $x=1$ is the only integer solution?
Related question: If [in a separate circumstance] I have
$$79x^4+40x^3-168x^2+96x-15=0,$$
which doesn’t factor algebraically into any linear factors, can I immediately conclude that the equation has no positive integer solutions? If not, what is the easiest way to prove that?
 A: Say you have $f(x) = a_nx^n  + \ldots a_1x = a_0$, a polynomial with integer coefficients; if it has a integer solution say $c$, then we have $c ( a_n c^{n-1} + \ldots + a_1) = a_0$ which implies that $c \mid a_0$. If $a_0$ is known then we just need to check whether the divisors  of $a_0$ satisfy the equation.
In the first question, we need to check whether all the divisors of $255$ satisfy the equation. In the second question, we need to check with all the divisors of  $15$. If none satisfy then there are  no integer roots.
A: There is the
rational root theorem
which tests the combined factors of $x^3$ and $x^0\space$
but the easiest way to see that
$\space (79x^3+159x^2-513x+255)=0\space$
has no integer solutions is to use Wolfram Alpha as shown
here.
Or use the "cubic formula" to find one root that can be used to reduce the cubic to a quadratic.
$$\text{Let}\quad a=79\qquad b=159\qquad c=-513\qquad d=255$$
\begin{align*}
x&=\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\
&+\sqrt[3]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\
&-\frac{b}{3a}   \end{align*}
This can often  be done in a spreadsheet where  it is common to put the  parentheses
$\bigg(\bigg)$
parts in separate cells  and have one or three other cells do the final computations. Sometimes there will be transient complex values from the radicals but there is robust software that can handle it, including Wolfram Alpha.
A third option is also tedious, a
Trigonometric solution for three real roots
here
or here.
It takes a long time to figure out and apply but it works better for finding all three roots if there are indeed three roots.
