Completely factor the polynomial $x^4-6x^3+4x^2+6x-5$ given one of its factors is x-1 .

A) $(x-1)(x+1)(x+5{)}^2$
B) $(x-9)(x+7)(x-1{)}^2$
C) $(x+1)(x-1)(x-5{)}^2$
D) $(x-5)(x+1)(x-1{)}^2$
E) $(x-1)(x+9)(x-7{)}^2$

We can use polynomial long division or synthetic division to divide the given polynomial by x-1. Using synthetic division, we get:

\begin{array}{c|ccccc}
1 & 1 & -6 & 4 & 6 & -5 \\
& \downarrow & 1 & -5 & -1 & 7 \\ \hline
& 1 & -5 & -1 & 5 & 2 \\
\end{array}

Therefore, we have:

$$x^4-6x^3+4x^2+6x-5=(x-1)(x^3-5x^2-x+5)$$

Now we can factor the cubic polynomial $x^3-5x^2-x+5$ using various methods (such as grouping or polynomial long division). One possible factorization is:

$$x^3-5x^2-x+5=(x-5)(x^2+1)$$

Combining this with the previous factorization, we get:

$$x^4-6x^3+4x^2+6x-5=(x-1)(x-5)(x^2+1)$$

And this is the final factorization. Option D matches this result, so it is the best choice.