1. TRINOMIALS
x2 + bx + c

2. When factoring trinomials, we look for a common factor before attempting any other technique. After this is done, the technique that we employ is determined by the number of terms in the polynomial.
Number of terms
Factoring Technique 2
Difference of 2 Squares 3
Trinomial
4 or 6
Grouping
This means that when we count three terms and after checking for a common factor, we try to factor using the trinomial method. In fact there are two versions of this method – a short cut method that can be used in certain situations and a longer, more drawn out method that can be used in all situations.

3. The method we will learn first is the short cut method which is a method we can apply to the simpler trinomials. These trinomials are distinctive due to the fact that the coefficient of the first term is always 1.
1x2 + bx + c
If that coefficient is anything other than 1, the short cut method will not apply and we must revert to the next method that we will learn. For now, we are learning the short cut method and therefore we will only use examples that have the variable 1.
First of all, it must be made clear that NOT ALL trinomials can be factored. Some cannot be factored because the numbers don’t work out and others can’t be factored because the variables don’t work out.
Our first examples will have the appropriate variables so we need only focus on the numbers.

4. Example 1: x2 + 11x + 24
When factoring these trinomials the factors will be two binomials: (x )(x )
We know that the first terms of each binomial must be x because the first term of the trinomial is x2 and x  x = x2. The challenge is to find the last term of each binomial. They must be chosen so that they will cause the coefficient of the middle term and the last term of the trinomial to work out. (That’s 11 and 24 in this case.)
The last terms of each trinomial must be chosen on this basis:
Their sum equals the middle term of the trinomial. +
= 11
Their product of those same numbers equals the last term of the trinomial. 
= 24

5. To determine these, it is easier to consider all of the limited number of factors that gives the last term and from that group of factors chose the pair that also gives the sum.
Factors of 24:
Example 1: x2 + 11x + 24 1 24 2 12 3 8 4 6
Upon inspecting the four pairs of factors, we can see they give us four distinct sums. 1 24
SUM = 25 2 12
SUM = 14 3 8
SUM = 11 4 6
SUM = 10 1 24
SUM = 25 2 12
SUM = 14 3 8
SUM = 11 4 6
SUM = 10 1 24
SUM = 25 2 12
SUM = 14 3 8
SUM = 11 4 6
SUM = 10 1 24
SUM = 25 2 12
SUM = 14 3 8
SUM = 11 4 6
SUM = 10 1 24
SUM = 25 2 12
SUM = 14 3 8
SUM = 11 4 6
SUM = 10
It is the factors 3 and 8 that produce a sum of 11 AND a product of 24 so they must be the last terms of each binomial.
(x + 3)(x + 8)

6. If we multiply these factors using FOIL, we get the polynomial that we started with.
(x)(x) = x2
(x)(8) = 8x
(x + 3)(x + 8)
(3)(x) = 3x
= x2 + 8x + 3x + 24
(3)(8) = 24
As we look at the 4 terms above, it becomes apparent why the sum of the last terms in each binomial must be equal to the middle term of the trinomial.
(x + 3)(x + 8)
Factors of 28: 1 28 2 14 4 7
Factors of 28: 1 28 2 14 4 7
= x2 + 8x + 3x + 24
= x2 + 11x + 24
SUM = 16
Example 2: a2 + 16a + 28
a  a = a2 so they are the first terms of each binomial and the factors 2 and 14 make a sum of 16 so the are the last terms of each binomial.
a2 + 16a + 28
= (a + 2)(a + 14)
= (a + 2)(a + 14)

7. Sometimes there is only 1 pair of factors to consider.
Factors of 1: 1
Factors of 1: 1
Example 3: y2 + 2y + 1
SUM = 2
y2 + 2y + 1
Sometimes there is only 1 pair of factors to consider.
= (y + 1)(y + 1)
= (y + 1)(y + 1)
Factors of 1: 1
Example 4: m2 + 3m + 1
SUM = 3
In this example the factors available do not make a sum of 3 which means that the trinomial can’t be factored.
Factors of 120: 1
120 2 60 3 40 4 30 5 24 6 20 8 15 10 12 1
120 2 60 3 40 4 30 5 24 6 20 8 15 10 12
Example 5: p2 + 23p + 120
In this example there are many pairs of factors to consider. Most examples will have fewer than these. The trick is in being able to quickly find all of the factors of c.
SUM = 23
p2 + 23p + 120
= (p + 8)(p + 15)
= (p + 8)(p + 15)

8. Example 6: x2 + 5x + 6 Factors of 6: Factors of 6: = (x + 2)(x + 3)1 6 2 3
Factors of 6: 1 6 2 3
= (x + 2)(x + 3)
SUM = 5
In each of the preceding examples the signs of the terms in the trinomials were always positive. Now we will observe examples where the signs can be negative.
Example 7: x2 + 5x - 6
Factors of -6:
= (x - 1)(x + 6) -1 +6 -2 +3 -1 +6 -2 +3
SUM = 5
When looking for the factors of a negative number, one must be positive and the other negative. If at the same time their sum is positive, then the factor that is bigger must be the positive one.

9. REVIEW OF RULES FOR SIGNS
Sign of bigger number
(+) + (+) = (+)
(+) + (-) =
(-) + (+) =
(-) + (-) = (-)
( )
ADDITION
(+)(+) = (+)
(+)(-) = (-)
(-)(+) = (-)
(-)(-) = (+)
MULTIPLICATION
Example 8: x2 - 5x - 6
Factors of -6:
= (x + 1)(x - 6) +1 -6 +2 -3
SUM = -5
When both the product and sum are negative, the factors have opposite signs but this time the bigger factor will be negative.

10. Example 9: x2 - 5x + 6 Factors of 6: = (x - 2)(x - 3) SUM = -5-1 -6 -2 -3
SUM = -5
When looking for factors of a positive number when the sum is negative, both factors will be negative.
Factors of -36:
Example 10: x2 - 5x - 36 1
-36 2
-18 3
-12 4 -9 6 -6 1
-36 2
-18 3
-12 4 -9 6 -6
= (x + 4)(x - 9)
SUM = -5