I'm adding a second answer to this post not because I think it is different in a fundamental way, but because it looks very different and feels very different in the execution. The title in the OP asks for an "easy way" to factorize, given that you know one of the factors already. One handy method is synthetic division, which works even with complex roots: one divides $z^4+3z^2-6z+10$ by $(z-(1+i))$ as follows:
$$
\begin{array}{r r r r r r}
\begin{array}{r |} 1+i\\ \hline \end{array}
& 1 & 0 & 3 & -6 & 10 \\
%\hline
& & 1+i & 2i & 1+5i & -10\\
\hline
& 1 & 1+i& 3+2i & -5+5i &\begin{array}{| r}0 \\ \hline \end{array} \\
\end{array}
$$
If you aren't familiar with the synthetic division algorithm: We work left-to-right, writing the sum of each column below the horizontal line, then multiplying that sum by the number in the top-left box and placing it one cell up and to the right.
The $0$ in the bottom-right cell confirms that $1+i$ is indeed a root of the polynomial, and the rest of the entries in the bottom row tell us that the quotient is $z^3 + (1+i)z^2 + (3+2i)z + (-5+5i)$ (the same as you found in your original solution). Now we have to factor this cubic.
At this point, we invoke the knowledge that complex roots of a polynomial with real coefficients come in conjugate pairs, so $1-i$ must be a root as well. We again use synthetic division on the quotient:
$$
\begin{array}{r r r r r }
\begin{array}{r |} 1-i\\ \hline \end{array}
& 1 & 1+i & 3+2i & -5+5i \\
%\hline
& & 1-i & 2-2i & 5-5i \\
\hline
& 1 & 2& 5 &\begin{array}{| r}0 \\ \hline \end{array} \\
\end{array}
$$
Again the $0$ in the bottom right corner confirms that $1-i$ is indeed a root, and the remaining coefficients in the bottom row tell us that the quotient is $z^2+2z+5$. Now use the quadratic formula to finish it.