How do you factor $2{{x}^{2}}-11x+6$?

Hint: For this problem we need to calculate the factors of the given equation. We can observe that the given equation is a quadratic equation which is in the form of $a{{x}^{2}}+bx+c$. We will compare both the equations to get the values of $a$, $b$, $c$. After getting the values of $a$, $b$, $c$ we will split the middle term $bx$ as ${{b}_{1}}x+{{b}_{2}}x$ such that ${{b}_{1}}{{b}_{2}}=ac$. Now we will take appropriate terms as common from the obtained equation and simplify the equation to get the required result.

Complete step by step solution:
Given that, $2{{x}^{2}}-11x+6$.
Comparing the above equation with the standard form of the quadratic equation which is $a{{x}^{2}}+bx+c$, then we will get
$a=2$, $b=-11$, $c=6$.
Now the value of $ac$ will be $ac=2\times 6=12$. We can write the factors of the value $12$ as $1$, $2$, $3$, $4$, $6$, $12$. From the above factors we can’t find any factors such that their sum is $-11$ and product is $12$. So, the equation can’t be factored.

Note: We can check whether the given equation is factored or not by calculating the value of ${{b}^{2}}-4ac$. In the given equation we have the values $a=2$, $b=-11$, $c=6$. From these values, the value of ${{b}^{2}}-4ac$ will be
$\begin{align}
  & \Rightarrow {{b}^{2}}-4ac={{\left( -11 \right)}^{2}}-4\left( 2 \right)\left( 6 \right) \\
 & \Rightarrow {{b}^{2}}-4ac=121-48 \\
 & \Rightarrow {{b}^{2}}-4ac=73 \\
\end{align}$
Here we have the value of ${{b}^{2}}-4ac$ as $73$ which is not a perfect square, so we can’t factorize the given equation. We can only factorize the equation when the value of ${{b}^{2}}-4ac$ is perfect square.