How do you factor the trinomial \[20{{x}^{2}}+22x+6\]?

Hint: In this problem, we have to factor the given trinomial. We can first take the common multiple 2 out of the equation. We can then factor the resulting equation. In the equation, we can split the middle term as their addition should be equal to the middle term itself and their product should be equal to the product of coefficients of \[{{x}^{2}}\] and constant. We can then separate it to find the factors for the given trinomial.

Complete step by step solution:
We know that the given trinomial is,
 \[20{{x}^{2}}+22x+6\]
 We can now take the common multiple 2 out of the equation, we get
\[\Rightarrow 2\left( 10{{x}^{2}}+11x+3 \right)\] …… (1)
We can now find the factor for the equation (1).
In the equation, we can split the middle term 11 as their addition should be equal to the middle term itself and their product should be equal to the product of coefficient of \[{{x}^{2}}\] and constant term.
\[\begin{align}
  & \Rightarrow 5+6=11 \\
 & \Rightarrow 5\times 6=30 \\
\end{align}\]
We can now write the equation (1) as,
\[\Rightarrow 2\left( 10{{x}^{2}}+5x+6x+3 \right)\]
We can now separate the terms and take the common terms, we get
\[\begin{align}
  & \Rightarrow 2\left( 5x\left( 2x+1 \right)+3\left( 2x+1 \right) \right) \\
 & \Rightarrow 2\left( 5x+3 \right)\left( 2x+1 \right) \\
\end{align}\]
Therefore, the factors of the trinomial \[20{{x}^{2}}+22x+6\] is \[2\left( 5x+3 \right)\left( 2x+1 \right)\].

Note: Students make mistakes while finding the factors, Here we have alternate approaches for factorisation which are algebraic identity, quadratic formula. So at the time of solving these types of problems we should know which method will be suitable for solving the problem easily.