Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+20$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-12$$ $$ab=1\times 20=20$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $20$.
$$-1,-20$$ $$-2,-10$$ $$-4,-5$$
Calculate the sum for each pair.
$$-1-20=-21$$ $$-2-10=-12$$ $$-4-5=-9$$
The solution is the pair that gives sum $-12$.
$$a=-10$$ $$b=-2$$
Rewrite $x^{2}-12x+20$ as $\left(x^{2}-10x\right)+\left(-2x+20\right)$.
$$\left(x^{2}-10x\right)+\left(-2x+20\right)$$
Factor out $x$ in the first and $-2$ in the second group.
$$x\left(x-10\right)-2\left(x-10\right)$$
Factor out common term $x-10$ by using distributive property.
$$\left(x-10\right)\left(x-2\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}-12x+20=0$$
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 $x^2+Bx+C=0$.
$$x ^ 2 -12x +20 = 0$$
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 $rs = C$
$$r + s = 12 $$ $$ rs = 20$$
Two numbers $r$ and $s$ sum up to $12$ exactly when the average of the two numbers is $\frac{1}{2}*12 = 6$. 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 $u$.
$$r = 6 - u$$ $$s = 6 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 20$
$$(6 - u) (6 + u) = 20$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$36 - u^2 = 20$$
Simplify the expression by subtracting $36$ on both sides
$$-u^2 = 20-36 = -16$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 16$$ $$u = \pm\sqrt{16} = \pm 4 $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
