How do you factor $8{{x}^{2}}+28x+12$ completely?

Hint: In this question we have been asked to factorize the given quadratic expression $8{{x}^{2}}+28x+12$ . For doing that we will have a quadratic expression to factorize for that we will use a traditional method. In traditional methods we need to write the coefficient of $x$ as the sum of the factors of the product of the coefficient of ${{x}^{2}}$ and the constant term.

Complete step by step solution:
Now considering from the question here we need to factorize the given quadratic expression $8{{x}^{2}}+28x+12$ .
We will first take the number four as common from the whole expression after that we will have $\Rightarrow 8{{x}^{2}}+28x+12=4\left( 2{{x}^{2}}+7x+3 \right)$ .
From the basic concepts we know that for factorization of a quadratic expression we will perform the traditional method. In traditional methods we need to write the coefficient of $x$ as the sum of the factors of the product of the coefficient of ${{x}^{2}}$ and the constant term.
So we need to write seven as the sum of factors of $2\times 3=6$
We can simply write this expression as $\Rightarrow 4\left( 2{{x}^{2}}+7x+3 \right)=4\left( 2{{x}^{2}}+6x+x+3 \right)$
We can further simplify and write this expression as
$\begin{align}
  & \Rightarrow 4\left( 2{{x}^{2}}+6x+x+3 \right)=4\left( 2x\left( x+3 \right)+1\left( x+3 \right) \right) \\
 & \Rightarrow 4\left( 2x+1 \right)\left( x+3 \right) \\
\end{align}$
Therefore we can conclude that the factors of the given expression $8{{x}^{2}}+28x+12$ are $4,\left( 2x+1 \right),\left( x+3 \right)$.

Note: While answering this question we should be sure with our concept that we apply and calculations we perform during the process of solution. Similarly this question can also be answered by using the formulae for finding the roots of any quadratic expression in the form of $a{{x}^{2}}+bx+c=0$ is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. If ${{x}_{1}},{{x}_{2}}$ are roots of this expression then the factors will be $\left( x-{{x}_{1}} \right),\left( x-{{x}_{2}} \right)$ .
For the expression $8{{x}^{2}}+28x+12$ the formulae can be applied and we will have $\begin{align}
  & x=\dfrac{-28\pm \sqrt{{{\left( 28 \right)}^{2}}-4\left( 8 \right)\left( 12 \right)}}{2\left( 8 \right)} \\
 & \Rightarrow x=\dfrac{-28\pm \sqrt{784-384}}{16} \\
 & \Rightarrow x=\dfrac{-28\pm \sqrt{400}}{16} \\
 & \Rightarrow x=\dfrac{-28\pm 20}{16} \\
 & \Rightarrow x=\dfrac{-8}{16},\dfrac{-48}{16}=\dfrac{-1}{2},-3 \\
\end{align}$
So now we will have $4,\left( 2x+1 \right)\text{and}\left( x+3 \right)$ as the factors of the expression.