﻿ Factoring Trinomials (coefficient of x^2 term is 1, constant term is negative)
FACTORING TRINOMIALS OF THE FORM $\,x^2 + bx + c\,$, WHERE $\,c\lt 0$
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!

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: $$\,\cssId{s11}{x^2 + bx + c} \ \ \cssId{s12}{=\ \ x^2 + (\overset{=\ b}{\overbrace{f+g}})x + \overset{=\ c}{\overbrace{\ fg\ }}} \ \ \cssId{s13}{=\ \ (x + f)(x + g)}$$

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.

For example, to mentally add $\,(-5) + 3\,$, in your head you would compute $\,5 - 3\,$,

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, $$\,\cssId{s64}{x^2 + 5x - 6} \ \ \cssId{s65}{=\ \ x^2 + (\overset{=\ 5}{\overbrace{6+(-1)}})x + \overset{=\ -6}{\overbrace{\ 6\cdot (-1)\ }}} \ \ \cssId{s66}{=\ \ (x + 6)(x - 1)}$$ Check: $\cssId{s68}{(x+6)(x-1)} \cssId{s69}{= x^2 - x + 6x - 6} \cssId{s70}{= x^2 + 5x - 6}$
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, $$\,\cssId{s85}{x^2 - 5x - 6} \ \ \cssId{s86}{=\ \ x^2 + (\overset{=\ -5}{\overbrace{(-6)+1}})x + \overset{=\ -6}{\overbrace{\ (-6)\cdot 1\ }}} \ \ \cssId{s87}{=\ \ (x - 6)(x + 1)}$$ Check: $\cssId{s89}{(x-6)(x+1)} \cssId{s90}{= x^2 + x - 6x - 6} \cssId{s91}{= x^2 - 5x - 6}$
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.
Master the ideas from this section

When you're done practicing, move on to:
Factoring Trinomials, All Mixed Up

CONCEPT QUESTIONS EXERCISE: