
"We will be examining quadratic trinomials of the form ax^{2} + bx + c."
Part 1: Trinomials with a = 1 (ax^{2} + bx + c) 

Method by Trial and Error:
Factor: x^{2}  x  12 
Step 1: The first step when factoring is to always check for a common factor among the terms. In this situation, there is no common factor. 
Nope! 
Step 2: To get the leading term of x^{2}, each first term of the factors will need to be x. 
(x )(x ) 
Step 3: The product of the last terms must be 12. Unfortunately, there are a variety of ways to arrive at this product of 12.

+12 and 1
12 and +1
+6 and  2
6 and +2
+4 and 3
4 and +3

ALL of these pairings
yield a product of 12.
But only ONE of them
is the correct choice!


Possible answers?
(x + 12)(x  1)
(x  12)(x + 1)
(x + 6)(x  2)
(x  6)(x + 2)
(x + 4)(x  3)
(x  4)(x + 3)

Step 4: Only one of the possible answers shown above will be the correct factoring for this problem. Which choice, when multiplied, will create the correct "middle term" needed for this example?
(The needed middle term for this problem is "x".) 
(x + 12)(x  1) middle term +11x
(x  12)(x + 1) middle term 11x
(x + 6)(x  2) middle term +4x
(x  6)(x + 2) middle term 4x
(x + 4)(x  3) middle term +x
(x  4)(x + 3) middle term x

Step 5: Write the answer. 
x^{2}  x  12 = (x  4)(x + 3) 
Yes, there is a shorter way to find the answer!
If the coefficient of x^{2} is 1,
then x^{2} + bx + c = (x + m)(x + n)
where m and n multiply to give c
and m and n add to give b.


In plain English:
If the leading coefficient is 1,
you need to find
"numbers that multiply to the last term
and add to the middle term's coefficient?"
Be careful of the signs of the numbers!
Part 2: Trinomials with a ≠ 1 (ax^{2} + bx + c) 
Unfortunately, when the leading coefficient is a number other than one, the number of possible answers increases, making the search for the correct answer more difficult.
Method by Trial and Error:
Factor: 3x^{2}  7x  6 
Step 1: The first step when factoring is to always check for a common factor among the terms. In this example, there is no common factor. 
Nope! 
Step 2: Consider all of the possible factors of the leading coefficient. In this example, there is only one way to arrive at 3x^{2}: the first terms of the factors will need to be 3x and x. 
(3x )(x ) 
Step 3: The product of the last terms must be 6. Unfortunately, there are several ways to arrive at this product of 6.

+6 and 1
6 and +1
+3 and  2
3 and +2

ALL of these pairings
yield a product of 6.

Notice in this step how it was necessary to consider each of these pairings with BOTH arrangements of 3x and x. Be sure you cover ALL of the possibilities. 
Possible answers?
(3x + 6)(x  1)
(3x  6)(x + 1)
(3x + 3)(x  2)
(3x  3)(x + 2)
(x + 6)(3x  1)
(x  6)(3x + 1)
(x + 3)(3x  2)
(x  3)(3x + 2)

Step 4: Only one of the possible answers shown above will be the correct factoring for this problem. Which choice, when multiplied, will create the correct "middle term" needed for this example?
(The needed middle term for this problem is "7x".) 
(3x + 6)(x  1) middle term +3x
(3x  6)(x + 1) middle term 3x
(3x + 3)(x  2) middle term 3x
(3x  3)(x + 2) middle term +3x
(x + 6)(3x  1) middle term +17x
(x  6)(3x + 1) middle term 17x
(x + 3)(3x  2) middle term +7x
(x  3)(3x + 2) middle term 7x

Step 5: Write the answer. 
3x^{2}  7x  6 = (x  3)(3x + 2) 
Sorry, there is no quick "short cut method" to find the answer!
Once you determine the possible factors for the leading term and for the constant term,
concentrate on finding the correct middle term.
Be careful of the signs!
Part 3: Not all trinomials can be factored. 
There are trinomials that simply do not comply with these factoring patterns. Not all trinomials can be factored using only integers (or over the set of integers).
Trinomials such as x² + 2x + 7 or 3x²  x  5 will not play by these factoring rules.
These trinomials, over the set of integers, are called prime polynomials.
In the Quadratics section, we will find additional procedures that will be helpful when attempting to factor polynomials that cannot be factored over the set of integers.


These first graphs show functions defined by expressions which are factorable over the set of integers. 


These second graphs show functions defined by expressions which are NOT factorable over the set of integers.These are prime polynomials over the set of integers. 
Factor: x^{2 } + 5x + 6 using Algebra Tiles

Place the x^{2} tile, 5 xtiles, and 6 1tiles in the grid. Be sure to keep straight lines inside the grid.

Fill the outside sections of the grid with xtiles and 1tiles that complete the pattern.


The graphing calculator can prove to be helpful for checking your answer when factoring over the set of integers. 

For help with factoring on your calculator,
Click Here!



NOTE: The reposting of materials (in part or whole) from this site to the Internet
is copyright violation
and is not considered "fair use" for educators. Please read the "Terms of Use". 
