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 $\,c\gt 0\,$.
That is, the constant term is positive.

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: $$\,x^2 + bx + c \ \ =\ \ x^2 + (\overset{=\ b}{\overbrace{f+g}})x + \overset{=\ c}{\overbrace{\ fg\ }} \ \ =\ \ (x + f)(x + g)$$ Since $\,c\,$ is positive in this exercise, both numbers will be positive, or both numbers will be negative.
(How can two numbers multiply to give a positive result? They must both be positive, or they must both be negative.)
That is, both numbers will have the same sign.

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

The sign of $\,b\,$ (the coefficient of the $\,x\,$ term) determines the common sign of your numbers.
If $\,b\gt 0\,$, then both numbers will be positive.
If $\,b\lt 0\,$, then both numbers will be negative.

These results are summarized below:

FACTORING TRINOMIALS OF THE FORM $\,x^2 + bx + c\,$,   $\,c\gt 0$
• Check that the coefficient of the square term is $\,1\,$.
• Check that the constant term ($\,c\,$) is positive.
• It's easier to do mental computations involving only positive numbers.
So, if $\,b\,$ (the coefficient of the $\,x\,$ term) is negative,
you'll initially ignore the minus sign.
That is, in the next step, notice that you're working with the absolute value of $\,b\,$,
which is positive.
• Find two numbers that ADD TO $\,|b|\,$ and MULTIPLY TO $\,c\,$.
• Now (and only now), you'll use the actual plus-or-minus sign of $\,b\,$.
If $\,b\gt 0\,$, then both numbers will be positive.
If $\,b\lt 0\,$, then both numbers will be negative.
• 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 positive?   Check!
Find two numbers that add to $\,5\,$ and multiply to $\,6\,$.
The numbers $\,2\,$ and $\,3\,$ work, since $\,2 + 3 = 5\,$ and $\,2\cdot 3 = 6\,$.
Since the coefficient of $\,x\,$ is positive, both numbers will be positive.
The desired numbers are $\,2\,$ and $\,3\,$.
Then, $$\,x^2 + 5x + 6 \ \ =\ \ x^2 + (\overset{=\ 5}{\overbrace{2+3}})x + \overset{=\ 6}{\overbrace{\ 2\cdot 3\ }} \ \ =\ \ (x + 2)(x + 3)$$ Check: $(x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6$
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 positive?   Check!
Find two numbers that add to $\,5\,$ and multiply to $\,6\,$.
(Notice that we initially ‘throw away’ the minus sign on the $\,5\,$,
because it's easier to do mental arithmetic with positive numbers.)
The numbers $\,2\,$ and $\,3\,$ work, since $\,2 + 3 = 5\,$ and $\,2\cdot 3 = 6\,$.
Since the coefficient of $\,x\,$ is negative, both numbers will be negative.
The desired numbers are $\,-2\,$ and $\,-3\,$.
Then, $$\,x^2 - 5x + 6 \ \ =\ \ x^2 + (\overset{=\ -5}{\overbrace{(-2)+(-3)}})x + \overset{=\ 6}{\overbrace{\ (-2)(-3)\ }} \ \ =\ \ (x - 2)(x - 3)$$ Check: $(x-2)(x-3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6$
Question: Factor:   $x^2 + x + 1$
Solution:
There are no integers that add to $\,1\,$ and multiply to $\,1\,$.
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
(coefficient of $\,x^2\,$ term is $\,1\,$, negative constant term)

CONCEPT QUESTIONS EXERCISE: