Factoring Trinomials of the Form $\,x^2 + bx + c\,,$ where $\,c\lt 0$
Before doing this exercise, you may want to study:
 Basic Concepts Involved in Factoring Trinomials
 Factoring Trinomials of the form $\,x^2 + bx + c\,,$ where $c\gt 0$
Here, you will practice factoring trinomials of the form $\,x^2 + bx + c\,,$ where $\,b\,$ and $\,c\,$ are integers, and $\,c\lt 0\,.$ That is, the constant term is negative.
Recall that the integers are: $$\ldots,\,3,\,2,\,1,\,0,\,1,\,2,\,3,\,\ldots$$
As discussed in Basic Concepts Involved in Factoring Trinomials, you must first find two numbers that add to $\,b\,$ and that multiply to $\,c\,,$ since then: $$ \begin{align} &\cssId{s11}{x^2 + bx + c}\cr &\qquad \cssId{s12}{=\ \ x^2 + (\overset{=\ b}{\overbrace{f+g}})x + \overset{=\ c}{\overbrace{\ fg\ }}}\cr\cr &\qquad \cssId{s13}{=\ \ (x + f)(x + g)} \end{align} $$
Since $\,c\,$ is negative in this exercise, one number will be positive, and the other will be negative. (How can two numbers multiply to give a negative result? One must be positive, and the other negative.) That is, the numbers will have different signs.
When you add numbers that have different signs, then in your head you actually do a subtraction problem. For example, to mentally add $\,(5) + 3\,,$ in your head you would compute $\,5  3\,,$ and then assign a negative sign to your answer.
Think of it this way: Start at zero on a number line. Walk $\,5\,$ units to the left, and $\,3\,$ units to the right. You end up at $\,2\,.$ You walked farther to the left than you did to the right, so your final answer is negative.
The sign of $\,b\,$ (the coefficient of the $\,x\,$ term) determines which number will be positive, and which will be negative. If $\,b\gt 0\,,$ then the bigger number (the one farthest from zero) will be positive. If $\,b\lt 0\,,$ then the bigger number (the one farthest from zero) will be negative. In other words, the biggest number takes the sign (plus or minus) of $\,b\,.$
These results are summarized below:
 Check that the coefficient of the square term is $\,1\,.$
 Check that the constant term ($\,c\,$) is negative.
 It's easier to do mental computations involving only positive numbers. So, you will initially ignore all minus signs and just work with the numbers $\,b\,$ and $\,c\,.$
 Find two numbers whose difference is $\,b\,$ and whose product is $\,c\,.$ That is, find two numbers that subtract to give $\,b\,$ and that multiply to give $\,c\,.$

Now (and only now), you'll use the actual plusorminus sign of $\,b\,.$
If $\,b\gt 0\,,$ then the bigger of your two numbers is positive; the other is negative.
If $\,b\lt 0\,,$ then the bigger of your two numbers is negative; the other is positive.
That is, the biggest number takes the sign (plus or minus) of $\,b\,.$
 Use these two numbers to factor the trinomial, as illustrated in the examples below.
 Be sure to check your answer using FOIL.
Examples
 Is the coefficient of the $\,x^2\,$ term equal to $\,1\,$? Check!
 Is the constant term negative? Check!
 Find two numbers whose difference is $\,5\,$ and whose product is $\,6\,.$ That is, find two numbers that subtract to give $\,5\,$ and that multiply to give $\,6\,.$ The numbers $\,1\,$ and $\,6\,$ work, since $\,6  1 = 5\,$ and $\,6\cdot 1 = 6\,.$
 Since the coefficient of $\,x\,$ is positive, the bigger number ($\,6\,$) will be positive, and the other will be negative. The desired numbers are therefore $\,6\,$ and $\,1\,.$
 Then, $$ \begin{align} &\cssId{s64}{x^2 + 5x  6}\cr &\qquad \cssId{s65}{=\ \ x^2 + (\overset{=\ 5}{\overbrace{6+(1)}})x + \overset{=\ 6}{\overbrace{\ 6\cdot (1)\ }}}\cr\cr &\qquad \cssId{s66}{=\ \ (x + 6)(x  1)} \end{align} $$
 Check: $$ \begin{align} &\cssId{s68}{(x+6)(x1)}\cr &\qquad \cssId{s69}{= x^2  x + 6x  6}\cr &\qquad \cssId{s70}{= x^2 + 5x  6} \end{align} $$
 Is the coefficient of the $\,x^2\,$ term equal to $\,1\,$? Check!
 Is the constant term negative? Check!
 Find two numbers whose difference is $\,5\,$ and whose product is $\,6\,.$ That is, find two numbers that subtract to give $\,5\,$ and that multiply to give $\,6\,.$ The numbers $\,1\,$ and $\,6\,$ work, since $\,6  1 = 5\,$ and $\,6\cdot 1 = 6\,.$
 Since the coefficient of $\,x\,$ is negative, the bigger number ($\,6\,$) will be negative, and the other will be positive. The desired numbers are therefore $\,6\,$ and $\,1\,.$
 Then, $$ \begin{align} &\cssId{s85}{x^2  5x  6}\cr &\qquad \cssId{s86}{=\ \ x^2 + (\overset{=\ 5}{\overbrace{(6)+1}})x + \overset{=\ 6}{\overbrace{\ (6)\cdot 1\ }}}\cr\cr &\qquad \cssId{s87}{=\ \ (x  6)(x + 1)} \end{align} $$
 Check: $$ \begin{align} &\cssId{s89}{(x6)(x+1)}\cr &\qquad \cssId{s90}{= x^2 + x  6x  6}\cr &\qquad \cssId{s91}{= x^2  5x  6} \end{align} $$