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 $ax^{2}+bx+c=0$.
$$3x^{2}-8x-10=0$$
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 $x=\frac{8±2\sqrt{46}}{6}$ when $±$ is plus. Add $8$ to $2\sqrt{46}$.
$$x=\frac{2\sqrt{46}+8}{6}$$
Divide $8+2\sqrt{46}$ by $6$.
$$x=\frac{\sqrt{46}+4}{3}$$
Now solve the equation $x=\frac{8±2\sqrt{46}}{6}$ when $±$ is minus. Subtract $2\sqrt{46}$ from $8$.
$$x=\frac{8-2\sqrt{46}}{6}$$
Divide $8-2\sqrt{46}$ by $6$.
$$x=\frac{4-\sqrt{46}}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{4+\sqrt{46}}{3}$ for $x_{1}$ and $\frac{4-\sqrt{46}}{3}$ 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 $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $3$
$$x ^ 2 -\frac{8}{3}x -\frac{10}{3} = 0$$
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 $rs = C$
$$r + s = \frac{8}{3} $$ $$ rs = -\frac{10}{3}$$
Two numbers $r$ and $s$ sum up to $\frac{8}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{8}{3} = \frac{4}{3}$. 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 $u$.
$$r = \frac{4}{3} - u$$ $$s = \frac{4}{3} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -\frac{10}{3}$
