Sect. 5.4
Factoring Trinomials
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Factoring x2 + bx + c
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Using a Trial Factor Table
Factoring ax2 + bx + c

Using a Grouping Factor Table
5.4
1
Remember
FOIL?
 (x + )(x + )
Examine the trinomial:
You can predict the
Binomial operation signs
5.4
 (x –
)(x – )
 (x +
)(x – )
 (x +
)(x –2 )
Factoring Trinomials with Leading
Coefficient=1
x2 + bx + c b and c are #s
1. Write the trinomial in descending
powers of one variable.
2. First - remove any common factor
(there may not be one)
3. (not in book) Write down the
binomial factor pair with middle
operators + or –
4. Create a “Trial Factor Table”:
4.1 List the possible factor pairs
of the 3rd term’s coefficient.
Factor
x  7 x  12
2
Therefore:
(x – 3 )(x – 4 )
factor pairs of 12 factor sum = -7
-1 x -12
-2 x -6
-3 x -4
4.2 List their corresponding sums.
5. Use the factor pair where the sum
of the factors is the coefficient of
the middle term.
5.4
-1 + -12 = -13 no
-2 + -6 = -8 no
-3 + -4 = -7 yes!
3
Using a Factor Table for x2 + bx + c
- Organized Trial & Error
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Let’s use x2 + 13x + 36 as an example
Factors must both be sums: (x + ?)(x + ?)
Pairs=c=36
Sum=b=13
1, 36
37
2, 18
20
3, 12
15
4, 9
13 ok quit!
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x2 + 13x + 36 = (x + 4)(x + 9)
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5.4
4
Factoring Practice:
x  9x  8
c8 | b9
( x  )( x  )
1,8
2
|
9 yes
( x  1)( x  8 )
y  20  9 y
2
y  9 y  20
2
c  20
| b  9
( y  )( y  )
 1,  20 |  21
( y  4 )( y  5 )
 2 ,  10 |  12
 4 ,  5 |  9 yes
5.4
5
Factoring Practice:
x  x  30 x
3
2
x ( x  x  30 )
c   30 | b   1
x ( x  )( x  )
1,  30 |  29
x ( x  5 )( x  6 )
2 ,  15 |  13
2
3 ,  10 |  7
5, 6
|  1 yes
5.4
6
Factoring Practice:
2 x  34 x  220
2
2 ( x  17 x  110 )
2
2 ( x  )( x  )
2 ( x  22 )( x  5 )
5.4
7
“Can’t Factor”?
x2 – x – 7
x2 + 3x – 42
When There are 2 Variables
x2 – 2xy – 48y2
5.4
8
Factor by Grouping
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8t3 + 2t2 – 12t – 3
2t2(4t + 1) – 3(4t + 1)
(4t + 1)(2t2 – 3)
4x3 – 6x2 – 6x + 9
2x2(2x – 3) – 3(2x – 3)
(2x – 3)(2x2 – 3)
y4 – 2y3 – 12y – 3
y3(y – 2) – 3(4y – 1)
Oops – not factorable via grouping
5.4
9
Factoring ax2 + bx + c (a not 1 or 0)
Factoring by Guessing
1. Write the trinomial in descending powers of one variable.
2. Factor out any greatest common factor (including 1, if that is
necessary to make the coefficient of the first term positive).
3. Determine the signs of the factors: (assume a is positive)
If c is + then both factors would need to have the sign of b
If c is – then the factors must have different signs
4. Try various combinations of the factors of the first terms and the
last terms until either:
You find a pair of binomial factors that work, or
Try all possible combinations but none work. (unfactorable)
5. Check the factorization by multiplication.
Let’s guess: 3p2 – 4p – 4 = (3p
5.4
)(p
)
10
Use Grouping to Factor
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16 + 24x + 5x2
5x2 + 24x + 16 (first rearrange)
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ac = 5(16)
= 2· 2· 2· 2· 5
1 · 80
2 · 40
4 · 20
b = 24
1 + 80 = 81
2 + 40 = 42
4 + 20 = 24
5x2 + 4x + 20x + 16
x(5x + 4) + 4(5x + 4)
(5x + 4)(x + 4) ta-da!
5.4
No
No
YES
11
The ac Grouping Method: ax2 + bx + c
Split bx into 2 Terms: Use a Grouping Factor Table
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Let’s use 3x2 – 10x – 8 as an example
ac = 3(-8) = -24
One factor is positive, the other negative and larger.
Pairs=ac=3(-8)
=-2·2·2·3
Sum=b=-10
1, -24
-23
2, -12
-10 quit!
3, -8
-5
4, -6
-2
3x2 – 10x
–8=
3x2 + 2x – 12x – 8 = split the middle
x(3x + 2) – 4(3x + 2) = do grouping
(3x + 2)(x – 4)
5.4
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ac Factoring Practice 1:
3x2 + 10x – 8
5.4
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ac Factoring Practice 2:
6x6 – 19x5 + 10x4
5.4
14
ac Factoring Practice 3:
6x4 – 116x3 – 80x2
5.4
15
What Next?

Section 5.5 –
Factoring Perfect
Squares & Differences
of Squares
5.4
16