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 $ax^{2}+bx+c=0$.
$$y^{2}-5y-25=0$$
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 $y=\frac{5±5\sqrt{5}}{2}$ when $±$ is plus. Add $5$ to $5\sqrt{5}$.
$$y=\frac{5\sqrt{5}+5}{2}$$
Now solve the equation $y=\frac{5±5\sqrt{5}}{2}$ when $±$ is minus. Subtract $5\sqrt{5}$ from $5$.
$$y=\frac{5-5\sqrt{5}}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5+5\sqrt{5}}{2}$ for $x_{1}$ and $\frac{5-5\sqrt{5}}{2}$ 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 $x^2+Bx+C=0$.
$$x ^ 2 -5x -25 = 0$$
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 $rs = C$
$$r + s = 5 $$ $$ rs = -25$$
Two numbers $r$ and $s$ sum up to $5$ exactly when the average of the two numbers is $\frac{1}{2}*5 = \frac{5}{2}$. 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 $u$.
$$r = \frac{5}{2} - u$$ $$s = \frac{5}{2} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -25$
$$(\frac{5}{2} - u) (\frac{5}{2} + u) = -25$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{25}{4} - u^2 = -25$$
Simplify the expression by subtracting $\frac{25}{4}$ on both sides
$$-u^2 = -25-\frac{25}{4} = -\frac{125}{4}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
