Download Factoring Quadratic Expressions Method

Factoring Quadratic Expressions
Method - Factoring Using Cross Products
This method is one of the most common methods currently used in math classes in Europe and
Asia. It is simply a method that organizes the factoring process of inspection. With inspection,
one is looking for the two binomial expressions that are the factors of the quadratic trinomial
expression. It is basically a “reversal” of the distributive property of multiplication in the
attempt to find the missing terms in the following: ( ? ± ? ) ( ? ± ? ) = ax 2  bx  c . For
those that refer to the acronym FOIL (First Outside Inside Last) as they apply the distributive
property to binomial multiplication, inspection involves finding the proper combination of
terms that satisfy this pattern:
First = ax 2
Last = c
Outside + Inside = bx
The Cross Product Method basically organizes this search. After the understanding of a few
associated patterns, the amount of time spent searching for the correct combination can be
minimized.
example: 12x2 + 10x – 12
Step 1 – Factor out the GCF
GCF is 2
Factoring the GCF: 2(6 x 2  5 x  6)
Step 2 – Cross Product number search
Step 3 – Assemble the factored form
Note:
A
C
B
D
A
C
B
D
AB = 6x2
CD = -6
AD + BC = 5x
First
Last
Outside + Inside
2(A+C)(B+D)
After you have removed the GCF in step 1, the correct binomial
factors of the quadratic expression will always have a GCF of 1
[As in, the GCF(A,C) = 1 ; GCF(B,D)=1]. Knowing this helps to
reduce the possibilities to try in your search.
Look at the work shown below that represents a student’s attempts to factor the expression
Remember: they are looking for a sum of cross products that is +5x
1st try
6x
-1
2nd try
6x
1
3rd try
6x
-6
x
x
x
6
sum 35x
-6
sum -35x
1
sum 0
4th try
6x
6
x
-1
sum 0
5th try
3x
-1
6th try
3x
6
2x
6
sum 16x
2x
-1
sum 9x
3x
-2
2x
3
sum 5x
Therefore, the factored form is 2(3x -2)(2x + 3)
How could they have saved time? Here are two shortcuts
-
The results of the 2nd try is the additive inverse of the results of the 1st try (same is true
between 3rd try and 4th try). Note the pattern with the additive inverse result: the
negative is simply assigned to the other factor. Use this shortcut when an additive
inverse sum is found
-
Secondly, note that the 3rd, 4th, 5th and 6th attempts had a GCF≠1 for the binomials
being formed. Therefore that combination could not be the correct one. There was no
need to attempt to determine the cross product sum for those attempts. Saves time
when you eliminate the need for that step.
Method - Factoring by Decomposition:
same example: 12x2 + 10x – 12
Step 1 – factor out the GCF
2(6 x 2  5 x  6)
Step 2 - Multiply the a and c terms in the quadratic expression
6 x -6 = -36
Step 2 – determine two numbers with a product equaling a X c with a sum of b
Sum (+5) Product (-36)
-6 , 6
4 , -9
+5
9, -4
The numbers are +9 and -4
Step 4 – decompose the middle term of the trinomial using the two numbers
6x2 -4x + 9x – 6
Step 5 – Factor out the GCF of the 1st two terms and of the 2nd two terms
6x2 – 2*3*x*x
4x - 2*2*x
GCF = 2x
9x – 3*3*x
6 - 2*3
GCF = 3
Step 6 – factor the resulting expression:
the original expression:
2x(3x -2) +3(3x – 2)
(3x-2)(2x+3). Then remember to return the GCF of
2(3x-2)(2x+3) is the factored form