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 $ax^{2}+bx+c=0$.
$$-6x^{2}+2x+5=0$$
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.
Now solve the equation $x=\frac{-2±2\sqrt{31}}{-12}$ when $±$ is plus. Add $-2$ to $2\sqrt{31}$.
$$x=\frac{2\sqrt{31}-2}{-12}$$
Divide $-2+2\sqrt{31}$ by $-12$.
$$x=\frac{1-\sqrt{31}}{6}$$
Now solve the equation $x=\frac{-2±2\sqrt{31}}{-12}$ when $±$ is minus. Subtract $2\sqrt{31}$ from $-2$.
$$x=\frac{-2\sqrt{31}-2}{-12}$$
Divide $-2-2\sqrt{31}$ by $-12$.
$$x=\frac{\sqrt{31}+1}{6}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{1-\sqrt{31}}{6}$ for $x_{1}$ and $\frac{1+\sqrt{31}}{6}$ for $x_{2}$.
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 $x^2+Bx+C=0$.
$$x ^ 2 -\frac{1}{3}x -\frac{5}{6} = 0$$
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 $rs = C$
$$r + s = \frac{1}{3} $$ $$ rs = -\frac{5}{6}$$
Two numbers $r$ and $s$ sum up to $\frac{1}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{1}{3} = \frac{1}{6}$. 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 $u$.
$$r = \frac{1}{6} - u$$ $$s = \frac{1}{6} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -\frac{5}{6}$
