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Standard Form
The standard form of any
2
ax
 bx  c
quadratic trinomial is
So, in 3x  4x  1...
2
a=3
b=-4
c=1
Now you try.
x  7x  2
2
a=
b=
c=
b=
c=
2x  x  5
2
a=
4x  x  2
2
a=
b=
c=
Factoring when a=1 and c > 0.
x
2
 8x  12

First list all the factor pairs of c.

Then find the factors with a sum of

These numbers are used to make the
factored expression.
x
 2x  6
1 , 12
2,6
3,4
b
Now you try.
x
2
 8x  15
x
Factors of c:
2
 10x  21
Factors of c:

Circle the factors of c with
the sum of b
Circle the factors of c with
the sum of b
Binomial Factors
(
)(
Binomial Factors
)
(
)(
)
Factoring when c >0 and b < 0.

c is positive and b is negative.
 Since a negative number times a negative
number produces a positive answer, we can
use the same method as before but…
 The binomial factors will have subtraction
instead of addition.
Let’s look at
x  13x  12
2
First list the factors of 12
1
12
We need 
a sum of -13
2
6
3
4
Make sure both values are negative!
x
 12x  1
Now you try.
2
1. x  7x  12
2. x  9x  14
2
3. x 13x  42
2
Factoring when c < 0.
We still look for the factors of c.
However, in this case, one factor should be
positive and the other negative in order to get a
negative value for c
Remember that the only way we can
multiply two numbers and come up
with a negative answer, is if one is
number is positive and the other is
negative!
Let’s look at
x  x  12
2
In this case, one factor
should be positive and the 1
other negative.
2

We need a sum of -1
x
 3x  4
+
3
12
6
-
4
Another Example x  3x  18
2
List the factors of 18.
We need a sum of 3

What factors and signs
will we use?
x
 3x  6
1
18
2
9
3
6

Now you try.
1.
2.
3.
4.
x
2
 3x  4
x
2
 x  20
x
2
 4 x  21
x
2
 10x  56
Prime Trinomials
Sometimes you will find a quadratic
trinomial that is not factorable.
You will know this when you
cannot get b from the list of
factors.
When you encounter this
write not factorable or
prime.
Here is an example…
x  3x  18
2
1
18
2
9
3
6
Since none of the pairs adds to 3,
this trinomial is prime.
Now you try.
x
2
 6x  4
x
2
 10x  39
x
2
 5x  7
factorable
factorable
factorable
prime
prime
prime
When a ≠ 1.
Instead of finding the factors of c:
Multiply a times c.
Then find the factors of this product.
1
70
2
7x  19x  10
a  c  70
2
35
5
14
7
10
We still determine the
factors that add to b.
So now we have
x
1
70
2
35
5
14
7
10
 5x  14
But we’re not finished yet….
Since we multiplied in the beginning,
we need to divide in the end.
Divide each constant by a.
 5  14 
x  x  
 7 
7 
Simplify, if possible.
 5 
x  x  2
 7 

Clear the fraction in each
binomial factor

7x  5x  2
2x  3x  9
2
Recall
2  9  18
• Multiply a times c.
• List factors. Look for sum of b
1
18
2
3
9
6
• Write 2 binomials using the
factors with sum of b 
x  6x  3
• Divide each constant by a.
 6 3
x  x  
 2  2 
• Simplify, if possible.
• Clear the fractions.


 3
x  3x  
 2 
x  32x  3
Now you try.
4x  4x  3
2
3x  5x 12
2
6x  23x  7
2
Sometimes there is a GCF.
If so, factor it out first.
4x  2x  30
22x  x 15
2
2
Then use the previous methods
to factor the trinomial




2x  6x  5

6 
5 
2x  x  

2 
2 

5 
2x  3x  

2 
2x  35x  2
Now you try.
1. 4 x 2  16x  12
2. 6x

2
 10x  6
45x  35x 10
2
Recall
59x 2  7x  2
First factor out the GCF.
5x  2x  9
Then factor the
remaining trinomial.


 2  9 
5x  x  
 9  9 
59x  2x 1


59x  2x 1


1.
6x  30x  36
2.
4x  14x  10
2
2