notice, to do a synthetic division, all terms have to be sorted first, and if any is missing, that simply means they have a coefficient of zero, but they're there.
[tex]\bf P(x)=x^3+x-10\implies P(x)=x^3\underline{+0x^2}+x-10
\\\\\\
\textit{since we know that 2 is a zero, then let's use it as divisor }
\begin{cases}
x=2\\
x-2=0
\end{cases}\\\\
-------------------------------\\\\
\begin{array}{r|rrrrrr}
2&&1&0&1&-10\\
&&&2&4&10\\
-&&-&-&-&-\\
&&1&2&5&0&\impliedby remainder
\end{array}
\\\\\\
\stackrel{given~zero}{(x-2)}(1x^2+2x+5)\implies (x-2)(x^2+2x+5)[/tex]
