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 $ax^{2}+bx+c=0$.
$$x^{2}-2x-4=0$$
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 $x=\frac{2±2\sqrt{5}}{2}$ when $±$ is plus. Add $2$ to $2\sqrt{5}$.
$$x=\frac{2\sqrt{5}+2}{2}$$
Divide $2+2\sqrt{5}$ by $2$.
$$x=\sqrt{5}+1$$
Now solve the equation $x=\frac{2±2\sqrt{5}}{2}$ when $±$ is minus. Subtract $2\sqrt{5}$ from $2$.
$$x=\frac{2-2\sqrt{5}}{2}$$
Divide $2-2\sqrt{5}$ by $2$.
$$x=1-\sqrt{5}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $1+\sqrt{5}$ for $x_{1}$ and $1-\sqrt{5}$ 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 $x^2+Bx+C=0$.
$$x ^ 2 -2x -4 = 0$$
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 $rs = C$
$$r + s = 2 $$ $$ rs = -4$$
Two numbers $r$ and $s$ sum up to $2$ exactly when the average of the two numbers is $\frac{1}{2}*2 = 1$. 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 $u$.
$$r = 1 - u$$ $$s = 1 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -4$
$$(1 - u) (1 + u) = -4$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$1 - u^2 = -4$$
Simplify the expression by subtracting $1$ on both sides
$$-u^2 = -4-1 = -5$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 5$$ $$u = \pm\sqrt{5} = \pm \sqrt{5} $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
