Factor out $3$. Polynomial $2x^{2}+x+2$ is not factored since it does not have any rational roots.
$$3\left(2x^{2}+x+2\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$6x^{2}+3x+6=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.
$$6x^{2}+3x+6$$
Steps Using Direct Factoring Method
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $6$
$$x ^ 2 +\frac{1}{2}x +1 = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = -\frac{1}{2} $$ $$ rs = 1$$
Two numbers $r$ and $s$ sum up to $-\frac{1}{2}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{1}{2} = -\frac{1}{4}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = -\frac{1}{4} - u$$ $$s = -\frac{1}{4} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 1$
$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = 1$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{1}{16} - u^2 = 1$$
Simplify the expression by subtracting $\frac{1}{16}$ on both sides
$$-u^2 = 1-\frac{1}{16} = \frac{15}{16}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
