6.3 Factoring Trinomials Whose Leading Coefficient Is Not One
FACTORING TRINOMIALS WHOSE LEADING COEFFICIENT IS NOT 1
In this section we will learn methods for factoring trinomials of the form
ax 2 + bx + c; a ≠ 0 or 1 :
i.e 9 x 2 + 12 x + 9 and 14 x 2 + 12 x − 23 .
There are two different methods used to factor these:
1.) Trial and Error: used when a and c are prime numbers.
2.) Factoring by Grouping: used when a and c are not prime
numbers.
Method 1: FACTORING BY TRIAL AND ERROR:
This method is nice to use when a and c are prime numbers.
Factor: 2 x 2 + 5 x + 3 by trial and error.
We list the possibilities to try since there are only a couple:
To get a 2x 2 , we can have ( 2 x _____ )( x _____ )
To get a 3, we can have ( 2 x + 1)( x + 3) , ( 2 x + 3)( x + 1) , ( 2 x − 3)( x − 1) ,
or ( 2 x − 1)( x − 3)
These four trinomials represent the only possibilities to get a 2x 2 and 3.
We choose the one that gives the correct middle term.
( 2 x + 1)( x + 3) , ( 2 x + 3)( x + 1) , ( 2 x − 3)( x − 1) , or ( 2 x − 1)( x − 3)
EXAMPLE: Factor the following
a.) 2 x 2 + 7 x + 5
1
b.) 5 y 2 − 17 y + 6
Method 2: FACTORING BY GROUPING:
To factor ax 2 + bx + c; a ≠ 0 or 1 .
1.) Find two numbers whose product is a ⋅ c . Make a complete list of these
two numbers.
2.) Then find the two numbers from above that will add together to get b .
3.) Rewrite the middle term bx using the numbers you found in 2. This will
give you 4 terms.
4.) Factor the 4 terms by grouping.
EXAMPLE: Use factoring by grouping to factor 6 x 2 + 17 x + 12.
2.) Find the sum of each pair.
1.) We need to list all the numbers
whose product
is 6 ⋅ 12 = 72. 72 = 23 ⋅ 32
1
72
-1 -72
1 +72=73
-1+(-72)= -73
2
36
-2 -36
2+36=38
-2+(-36)= -38
3
12
-3 -12
3+12=15
-3+(-12)= -15
4
18
-4 -18
4+18=32
-4+(-18)= -32
6
12
-6 -12
6+12=18
-6+(-12)= -18
8
9
-8 -9
8+9=17
-8+(-9)= -17
We want the pair that gives 17.
8 and 9 are the numbers
3.) We chose 8 and 9 because 8+9=17.
Or 8 x + 9 x = 17 x . We replace 17x with 8 x + 9 x in 6 x 2 + 17 x + 12.
6 x 2 + 17 x + 12 = 6 x 2 + 8 x + 9 x + 12
2
now factor by grouping:
( 6 x2 + 8x ) + ( 9 x + 12 ) = 2 x ( 3x + 4 ) + 3( 3x + 4 ) = ( 2 x + 3)( 3x + 4 )
TRIAL AND ERROR
Let’s try to factor 6 x 2 + 17 x + 12 another way.
Step1: We must find two factors so that the product is the first term, 6 x 2 .
Two possibilities: 3x & 2 x or x & 6 x
6 x 2 + 17 x + 12 = ( 3 x + ? )( 2 x + ? )
6 x 2 + 17 x + 12 = ( x + ? )( 6 x + ? )
Step 2: We must find two factors so that the product is the last term,12.
The possibilities are: 1 & 12; 2 & 6; 3 & 4. We neglect negatives
since the middle
term is positive.
6 x 2 + 17 x + 12 = ( 3 x + 1)( 2 x + 12 )
6 x 2 + 17 x + 12 = ( x + 1)( 6 x + 12 )
6 x 2 + 17 x + 12 = ( 3 x + 2 )( 2 x + 6 )
6 x 2 + 17 x + 12 = ( x + 2 )( 6 x + 6 )
6 x 2 + 17 x + 12 = ( 3x + 3)( 2 x + 4 )
6 x 2 + 17 x + 12 = ( x + 3)( 6 x + 4 )
6 x 2 + 17 x + 12 = ( 3 x + 12 )( 2 x + 1)
6 x 2 + 17 x + 12 = ( 3 x + 6 )( 2 x + 2 )
6 x 2 + 17 x + 12 = ( 3x + 4 )( 2 x + 3)
6 x 2 + 17 x + 12 = ( x + 12 )( 6 x + 1)
6 x 2 + 17 x + 12 = ( x + 6 )( 6 x + 2 )
6 x 2 + 17 x + 12 = ( x + 4 )( 6 x + 3)
Step 3: We have a list of all the possible factorizations. We need the one
that gives us the correct middle term, 17 x.
( 3x + 1)( 2 x + 12 ) = 6 x 2 + 36 x + 4 x + 12
( 3x + 12 )( 2 x + 1) = 6 x 2 + 3x + 24 x + 12
( 3x + 2 )( 2 x + 6 ) = 6 x 2 + 18 x + 4 x + 12
( 3x + 6 )( 2 x + 2 ) = 6 x 2 + 6 x + 12 x + 12
( 3x + 3)( 2 x + 4 ) = 6 x 2 + 12 x + 6 x + 12
( 3x + 4 )( 2 x + 3) = 6 x 2 + 9 x + 8 x + 12
( x + 1)( 6 x + 12 ) = 6 x 2 + 12 x + 6 x + 12
( x + 12 )( 6 x + 1) = 6 x 2 + x + 72 x + 12
( x + 2 )( 6 x + 6 ) = 6 x 2 + 6 x + 12 x + 12
( x + 6 )( 6 x + 2 ) = 6 x 2 + 2 x + 36 x + 12
( x + 3)( 6 x + 4 ) = 6 x 2 + 4 x + 18 x + 12
( x + 4 )( 6 x + 3) = 6 x 2 + 3x + 24 x + 12
We look for the factorization that produces the correct middle term: 17x
3
( 3x + 4 )( 2 x + 3) = 6 x 2 + 9 x + 8 x + 12 = 6 x 2 + 17 x + 12
We conclude by writing the correct factorization:
6 x 2 + 17 x + 12 = ( 3 x + 4 )( 2 x + 3)
Both methods are good; however the trial and error method can take a
while if a and c have several factors.
EXAMPLE: Factor the following
a.) 3x 2 + 7 x + 2
b.) 6 y 2 + 7 y − 24
c.) 24 x 4 + 10 x3 − 4 x 2
4
d.) 16r 2 − 16r − 12
e.) 9r 2 − 12r + 4
f.) 9 x 2 + 3x + 2
5
g.) 3x 2 + 5 xy − 2 y 2
h) 4 x 2 − 20 x + 25
6