Factor the expression by grouping. First, the expression needs to be rewritten as $r^{2}+ar+br+16$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-10$$ $$ab=1\times 16=16$$
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 $16$.
$$-1,-16$$ $$-2,-8$$ $$-4,-4$$
Calculate the sum for each pair.
$$-1-16=-17$$ $$-2-8=-10$$ $$-4-4=-8$$
The solution is the pair that gives sum $-10$.
$$a=-8$$ $$b=-2$$
Rewrite $r^{2}-10r+16$ as $\left(r^{2}-8r\right)+\left(-2r+16\right)$.
$$\left(r^{2}-8r\right)+\left(-2r+16\right)$$
Factor out $r$ in the first and $-2$ in the second group.
$$r\left(r-8\right)-2\left(r-8\right)$$
Factor out common term $r-8$ by using distributive property.
$$\left(r-8\right)\left(r-2\right)$$
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 $ax^{2}+bx+c=0$.
$$r^{2}-10r+16=0$$
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 $r=\frac{10±6}{2}$ when $±$ is plus. Add $10$ to $6$.
$$r=\frac{16}{2}$$
Divide $16$ by $2$.
$$r=8$$
Now solve the equation $r=\frac{10±6}{2}$ when $±$ is minus. Subtract $6$ from $10$.
$$r=\frac{4}{2}$$
Divide $4$ by $2$.
$$r=2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $8$ for $x_{1}$ and $2$ for $x_{2}$.
$$r^{2}-10r+16=\left(r-8\right)\left(r-2\right)$$
Steps Using Direct Factoring Method
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 $x^2+Bx+C=0$.
$$x ^ 2 -10x +16 = 0$$
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 $rs = C$
$$r + s = 10 $$ $$ rs = 16$$
Two numbers $r$ and $s$ sum up to $10$ exactly when the average of the two numbers is $\frac{1}{2}*10 = 5$. 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 $u$.
$$r = 5 - u$$ $$s = 5 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 16$
$$(5 - u) (5 + u) = 16$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$25 - u^2 = 16$$
Simplify the expression by subtracting $25$ on both sides
$$-u^2 = 16-25 = -9$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 9$$ $$u = \pm\sqrt{9} = \pm 3 $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
