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.
$$a+b=6$$ $$ab=1\left(-16\right)=-16$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-16$.
$$-1,16$$ $$-2,8$$ $$-4,4$$
Calculate the sum for each pair.
$$-1+16=15$$ $$-2+8=6$$ $$-4+4=0$$
The solution is the pair that gives sum $6$.
$$a=-2$$ $$b=8$$
Rewrite $x^{2}+6x-16$ 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 $ax^{2}+bx+c=0$.
$$x^{2}+6x-16=0$$
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.
$$x=\frac{-6±\sqrt{6^{2}-4\left(-16\right)}}{2}$$
Square $6$.
$$x=\frac{-6±\sqrt{36-4\left(-16\right)}}{2}$$
Multiply $-4$ times $-16$.
$$x=\frac{-6±\sqrt{36+64}}{2}$$
Add $36$ to $64$.
$$x=\frac{-6±\sqrt{100}}{2}$$
Take the square root of $100$.
$$x=\frac{-6±10}{2}$$
Now solve the equation $x=\frac{-6±10}{2}$ when $±$ is plus. Add $-6$ to $10$.
$$x=\frac{4}{2}$$
Divide $4$ by $2$.
$$x=2$$
Now solve the equation $x=\frac{-6±10}{2}$ when $±$ is minus. Subtract $10$ from $-6$.
$$x=-\frac{16}{2}$$
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 $p+q$.
$$x^{2}+6x-16=\left(x-2\right)\left(x+8\right)$$
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 $x^2+Bx+C=0$.
$$x ^ 2 +6x -16 = 0$$
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 $rs = C$
$$r + s = -6 $$ $$ rs = -16$$
Two numbers $r$ and $s$ sum up to $-6$ exactly when the average of the two numbers is $\frac{1}{2}*-6 = -3$. 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 $u$.
$$r = -3 - u$$ $$s = -3 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -16$
$$(-3 - u) (-3 + u) = -16$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$9 - u^2 = -16$$
Simplify the expression by subtracting $9$ on both sides
$$-u^2 = -16-9 = -25$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 25$$ $$u = \pm\sqrt{25} = \pm 5 $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
