Using the factor theorem factorise $f\left( x \right) = {x^2} - 5x + 6$ .
A) $\left( {x + 2} \right)\left( {x - 3} \right)$
B) $\left( {x - 2} \right)\left( {x - 3} \right)$
C) $\left( {x - 2} \right)\left( {x + 3} \right)$
D) $\left( {x + 2} \right)\left( {x + 3} \right)$

Hint: We will substitute different values for the unknown and then find the root of the equation. After that we will express it in the factor form and divide the given polynomial with the obtained factor and obtain another factor and express the given polynomial as a product of two factors. That will be the factorisation of the given polynomial.

Complete step-by-step answer:
Clearly if we put $x = 0$, the equation is not satisfied so it is not a root of the polynomial.
Then we will put $x = 1$ and then we get the value of the polynomial as $f\left( 1 \right) = 2$.
Now we will start substituting the values on words.
After this we will put $x = 2$ and then we get the value of the polynomial as $f\left( 2 \right) = 0$ .
Therefore, $x = 2$ is the factor of the polynomial $f\left( x \right) = {x^2} - 5x + 6$ .
The factor theorem states that if $x = a$ is a root of the polynomial $p\left( x \right)$ then we say that $\left( {x - a} \right)$ is a factor of the polynomial $p\left( x \right)$.
Therefore, in the given case we can say that $\left( {x - 2} \right)$ is a factor $f\left( x \right) = {x^2} - 5x + 6$ .
On dividing the polynomial $f\left( x \right) = {x^2} - 5x + 6$ by $\left( {x - 2} \right)$ we get the following:
$f\left( x \right) = {x^2} - 5x + 6 \Rightarrow f\left( x \right) = \left( {x - 2} \right)\left( {x - 3} \right)$
Thus, the factorisation of the given polynomial is:
${x^2} - 5x + 6 = \left( {x - 2} \right)\left( {x - 3} \right)$

Hence, the correct option is B.

Note: Polynomials are the algebraic expressions which consist of variables and coefficients. Variables are also sometimes called indeterminates. We can perform the arithmetic operations such as addition, subtraction, multiplication and also positive integer exponents for polynomial expressions but not division by variable.