Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+16$. 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 positive, $a$ and $b$ are both positive. 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 $10$.
$$a=2$$ $$b=8$$
Rewrite $x^{2}+10x+16$ as $\left(x^{2}+2x\right)+\left(8x+16\right)$.
$$\left(x^{2}+2x\right)+\left(8x+16\right)$$
Factor out $x$ in the first and $8$ in the second group.
$$x\left(x+2\right)+8\left(x+2\right)$$
Factor out common term $x+2$ by using distributive property.
$$\left(x+2\right)\left(x+8\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+10x+16=0$$
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.
$$x=\frac{-10±\sqrt{10^{2}-4\times 16}}{2}$$
Square $10$.
$$x=\frac{-10±\sqrt{100-4\times 16}}{2}$$
Multiply $-4$ times $16$.
$$x=\frac{-10±\sqrt{100-64}}{2}$$
Add $100$ to $-64$.
$$x=\frac{-10±\sqrt{36}}{2}$$
Take the square root of $36$.
$$x=\frac{-10±6}{2}$$
Now solve the equation $x=\frac{-10±6}{2}$ when $±$ is plus. Add $-10$ to $6$.
$$x=-\frac{4}{2}$$
Divide $-4$ by $2$.
$$x=-2$$
Now solve the equation $x=\frac{-10±6}{2}$ when $±$ is minus. Subtract $6$ from $-10$.
$$x=-\frac{16}{2}$$
Divide $-16$ by $2$.
$$x=-8$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-2$ for $x_{1}$ and $-8$ for $x_{2}$.
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
$$x^{2}+10x+16=\left(x+2\right)\left(x+8\right)$$
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 $s$.
