Verify whether the indicated numbers are the zeros of the given polynomial
$p\left( x \right)={{x}^{3}}-6{{x}^{2}}+11x-6,x=1,2,3$

Hint: Recall the definition of a zero of a polynomial. If ${{x}_{0}}$is a zero of a polynomial p(x), then we have $p\left( {{x}_{0}} \right)=0$. Hence determine at which point p(x) vanishes. Those points are the roots of p(x).

Complete step-by-step answer:
Alternatively, use synthetic division to determine whether the given points are the roots of the polynomial or not.

We know that if ${{x}_{0}}$ is a root of a polynomial p(x), then we have $p\left( {{x}_{0}} \right)=0$.
Checking x = 1
We have $p\left( x \right)={{x}^{3}}-6{{x}^{2}}+11x-6$
Substituting x = 1 in the p(x), we get
$p\left( 1 \right)={{1}^{3}}-6{{\left( 1 \right)}^{2}}+11-6=1-6+11-6=0$
Since p(1) = 0, we have x = 1 is a zero of the polynomial p(x)
Checking x =2
We have $p\left( x \right)={{x}^{3}}-6{{x}^{2}}+11x-6$
Substituting x = 2 in the p(x), we get
$p\left( 2 \right)={{2}^{3}}-6{{\left( 2 \right)}^{2}}+11\times 2-6=8-24+22-6=0$
Since p(2) = 0, we have x = 2 is a zero of the polynomial p(x)
Checking x= 3
We have $p\left( x \right)={{x}^{3}}-6{{x}^{2}}+11x-6$
Substituting x = 3 in the p(x), we get
$p\left( 3 \right)={{3}^{3}}-6{{\left( 3 \right)}^{2}}+11\times 3-6=27-54+33-6=0$
Since p(3) = 0, we have x = 3 is a zero of the polynomial p(x).
Hence x = 1,2,3 are the zeros of the given polynomial p(x).

Note: Alternative method: Synthetic division: Best method.
In this method, we start by writing coefficients of the polynomial in order from the highest degree to the constant term. If in between some degree terms are missing we set their coefficient as 0.
Hence $p\left( x \right)={{x}^{3}}-6{{x}^{2}}+11-6$ will be written as shown below

Now the point which has to be substituted(say x= 1) is written as follows


0 is placed below the first term


Now the terms under the same column are added. The sum is then multiplied with the root, and the product is written under the coefficient of the next term.
Hence, we have


Continuing in this way we have the following


Since the last sum is 0, we have $p\left( 1 \right)=0$. This method also tells you what the quotient will be when p(x) is divided by x-1. Here it will be $1\left( {{x}^{2}} \right)-5\left( x \right)+6={{x}^{2}}-5x+6$
Similarly creating tables for x = 2 and x = 3, we get


Since the last sum is 0, we have $p\left( 2 \right)=0$


Since the last sum is 0, we have $p\left( 3 \right)=0$
Hence x =1,2,3 are the roots of the given polynomial.