keeping in mind that complex solutions do not come all by their lonesome
they're always in pairs, so -10i has a sister, +10i, or its conjugate
drawing from that we could say that [tex]p(x)=0
\\ \quad \\
p(x)=
\begin{cases}
x=-10i\implies x+10i=0\implies &(x+10i)=0
\\ \quad \\
x=+10i\implies x-10i=0\implies &(x-10i)=0
\end{cases}
\\ \quad \\
thus
\\ \quad \\
p(x)=0\implies (x+10i)(x-10i)=0
\\ \quad \\
--------------\\
\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\\
--------------\\
then
\\ \quad \\
(x+10i)(x-10i)=0\implies [x^2-(10i)^2]=0
\\ \quad \\\
[x^2-(10^2\cdot i^2)]=0\implies [x^2-(100\cdot -1)]=0[/tex]
and surely you can take it from there :)
