(b) Show that $x-1$ is a factor of $P(x)=x^{3}-6 x^{2}+11 x-6$ and hence, determine all factors.

Step 1 ：First, we check if \(x-1\) is a factor of \(P(x) = x^3 - 6x^2 + 11x - 6\) by using the Remainder Theorem. We evaluate \(P(1)\):

Step 2 ：P(1) = (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0

Step 3 ：Since \(P(1) = 0\), we know that \(x-1\) is a factor of \(P(x)\).

Step 4 ：Now, we perform polynomial long division to find the other factor:

Step 5 ：\[\begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & -5x & +6 \\ \cline{2-5} x-1 & x^3 & -6x^2 & +11x & -6 \\ \cline{2-2} \multicolumn{2}{r}{x^3} & -x^2 \\ \cline{2-3} \multicolumn{2}{r}{0} & -5x^2 & +11x \\ \cline{3-3} \multicolumn{2}{r}{} & -5x^2 & +5x \\ \cline{3-4} \multicolumn{2}{r}{} & 0 & 6x & -6 \\ \cline{4-4} \multicolumn{2}{r}{} & & 6x & -6 \\ \cline{4-5} \multicolumn{2}{r}{} & & 0 & 0 \end{array}\]

Step 6 ：The quotient is \(x^2 - 5x + 6\), which can be factored further as \((x-2)(x-3)\).

Step 7 ：Thus, the complete factorization of \(P(x)\) is \(\boxed{(x-1)(x-2)(x-3)}\).