The Invariant Root

Since $p(2)=13$ and $p(10)=5$ , by the Remainder-Factor Theorem ,

$p(x)={(x-2)}\times{q(x)}+13={(x-10)}\times{h(x)}+5$

where $q$ and $h$ are polynomials with integer coefficients.

Let $\alpha$ be an integral root of $p(x)$ .

$\implies p(\alpha)=0$

$\implies {(\alpha-2)}\times{q(\alpha)}+13={(\alpha-10)}\times{h(\alpha)}+5=0$

$\implies q(\alpha)=\frac{-13}{\alpha-2}$

$\text{and}$

$h(\alpha)=\frac{-5}{\alpha-10}$

$\implies (\alpha-2)|13$ and $(\alpha-10)|5$

$\implies |\alpha|=\boxed{15}$

Let 'k' be the integral solution of the polynomial p(x). Then (x-k) will be one of the factor of p(x). Also since p(x) has integral coefficients, p(x) will be one of the form of : 1) p(x)= (x-k)(ax-b)(cx-d)(ex-f)(gx-h) when all the roots are real. Here all coefficients are integral.

2) p(x)= (x-k)(ax-b)(cx-d)(e x^{2} - f^{2} ) when two of the roots are complex numbers. Here f is a complex number with integral coefficients.

3) p(x)= (x-k)(a x^{2} - b^{2} )(c x^{2} - d^{2} ) when there is only one real root. Here b and d are complex numbers with integral coefficients.

Now p(2)=13 and p(10)=5 implies that (2-k) and (10-k) will divide 13 and 5 respectively which left us with only one choice of k and that will be k=15.

So k=15 will always be root of this polynomial.