Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-k^{2}-10k-16$$
Factor the expression by grouping. First, the expression needs to be rewritten as $-k^{2}+ak+bk-16$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-10$$ $$ab=-\left(-16\right)=16$$
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 $-10$.
$$a=-2$$ $$b=-8$$
Rewrite $-k^{2}-10k-16$ as $\left(-k^{2}-2k\right)+\left(-8k-16\right)$.
$$\left(-k^{2}-2k\right)+\left(-8k-16\right)$$
Factor out $k$ in the first and $8$ in the second group.
$$k\left(-k-2\right)+8\left(-k-2\right)$$
Factor out common term $-k-2$ by using distributive property.
$$\left(-k-2\right)\left(k+8\right)$$
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 $ax^{2}+bx+c=0$.
$$-k^{2}-10k-16=0$$
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.
