Answer
$f(x)=(x-1)(x-1)(x^{2}+9)=0$
when $x\in\left\{1\right\}$
Work Step by Step
We try to find rational zeros of $f(x)=x^{4}-2x^{3}+10x^{2}-18x+9$
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 3,\pm 9}{\pm 1}$
Testing with synthetic division, ... we try $x-1$
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1&-2 & 10 & -18 &9 \\\hline
&1 & -1 & 9 & -9 \\\hline
1&-1 & 9 &-9 & |\ \ 0 \end{array}$
$f(x)=(x-1)(x^{3}-x^{2}+9x-9)$
... factor in pairs
$x^{3}-x^{2}+9x-9=x^{2}(x-1)+9(x-1)$
$=(x-1)(x^{2}+9)$
... no further factorization, as $x^{2}+9$ has no real zeros.
$f(x)=(x-1)(x-1)(x^{2}+9)=0$
when $x\in\left\{1\right\}$
