# Factoring Trinomials of the Form $\,x^2 + bx + c\,,$ where $\,c\lt 0$

Before doing this exercise, you may want to study:

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.

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:

FACTORING TRINOMIALS OF THE FORM $\,x^2 + bx + c\,,$   $\,c\lt 0$
• 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 plus-or-minus 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.

## Examples

Question: Factor:   $x^2 + 5x - 6$
Solution: Thought process:
• 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)(x-1)}\cr &\qquad \cssId{s69}{= x^2 - x + 6x - 6}\cr &\qquad \cssId{s70}{= x^2 + 5x - 6} \end{align}
Question: Factor:   $x^2 - 5x - 6$
Solution: Thought process:
• 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}{(x-6)(x+1)}\cr &\qquad \cssId{s90}{= x^2 + x - 6x - 6}\cr &\qquad \cssId{s91}{= x^2 - 5x - 6} \end{align}
Question: Factor:   $x^2 + x - 1$
Solution: There are no integers whose difference and product are both $\,1\,.$ Thus, $\,x^2 + x - 1\,$ is not factorable over the integers.