Consider $x^{2}-2x-80$. Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-80$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-2$$ $$ab=1\left(-80\right)=-80$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-80$.
Rewrite $x^{2}-2x-80$ as $\left(x^{2}-10x\right)+\left(8x-80\right)$.
$$\left(x^{2}-10x\right)+\left(8x-80\right)$$
Factor out $x$ in the first and $8$ in the second group.
$$x\left(x-10\right)+8\left(x-10\right)$$
Factor out common term $x-10$ by using distributive property.
$$\left(x-10\right)\left(x+8\right)$$
Rewrite the complete factored expression.
$$2\left(x-10\right)\left(x+8\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}-4x-160=0$$
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.
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 $2$
$$x ^ 2 -2x -80 = 0$$
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 $rs = C$
$$r + s = 2 $$ $$ rs = -80$$
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 $rs = -80$
$$(1 - u) (1 + u) = -80$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$1 - u^2 = -80$$
Simplify the expression by subtracting $1$ on both sides
$$-u^2 = -80-1 = -81$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 81$$ $$u = \pm\sqrt{81} = \pm 9 $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
