How do you factor $2{{x}^{2}}+4x+6$?

Hint: We will first find the roots of the given quadratic expression by using the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Now the factors of the given expression are $\left( x-\alpha \right)$ and $\left( x-\beta \right)$ where $\alpha $ and $\beta $ are the roots of the given expression. Hence we have the roots of the expression.

Complete step by step solution:
The given expression is a quadratic expression of the form $a{{x}^{2}}+bx+c$ where a = 2, b = 4 and c = 6.
Now we want to find the factors of the given expression. Factors are nothing but polynomials which divide the given polynomial.
Now to factor the given expression we will first find the roots of the expression.
Roots of a quadratic expression is the value of x for which the expression is 0.
We know that for a quadratic expression of the form $a{{x}^{2}}+bx+c$ the roots are given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Hence the roots of the given quadratic are,
$\begin{align}
  & \Rightarrow \dfrac{-4\pm \sqrt{{{4}^{2}}-4\left( 2 \right)\left( 6 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow \dfrac{-4\pm \sqrt{16-48}}{4} \\
 & \Rightarrow \dfrac{-4\pm 32i}{4} \\
 & \Rightarrow -1\pm 8i \\
\end{align}$
Hence the roots of the expression are $-1-8i$ and $-1+8i$ .
Now we know that if $\alpha $ and $\beta $ are the roots of the expression then $\left( x-\alpha \right)$ and $\left( x-\beta \right)$ are the expression.
Hence we have $\left( x-\left( -1-8i \right) \right)$ and $\left( x-\left( -1+8i \right) \right)$ are the factors of the given expression.

Note: Now the roots of the expression can be real or complex. The nature of roots depends on the discriminant of quadratic which is defined as ${{b}^{2}}-4ac$ if the discriminant is greater than zero then we have real roots if it is zero then we have real and equal roots and if it is negative then the roots are complex.