1. Factoring Trinomials and Making Lists

2. For our discussions, trinomials (three-term) in x are polynomials
of the form ax2 + bx + c where a (≠ 0), b, and c are numbers.
Factoring Trinomials and Making Lists

3. For our discussions, trinomials (three-term) in x are polynomials
of the form ax2 + bx + c where a (≠ 0), b, and c are numbers.
In general, we have (#x + #)(#x + #)  ax2 + bx + c.
Factoring Trinomials and Making Lists

4. For our discussions, trinomials (three-term) in x are polynomials
of the form ax2 + bx + c where a (≠ 0), b, and c are numbers.
In general, we have (#x + #)(#x + #)  ax2 + bx + c.
For example,
(x + 2)(x + 1)  x2 + 3x + 2 with a = 1, b = 3, and c = 2.
Factoring Trinomials and Making Lists

5. For our discussions, trinomials (three-term) in x are polynomials
of the form ax2 + bx + c where a (≠ 0), b, and c are numbers.
In general, we have (#x + #)(#x + #)  ax2 + bx + c.
For example,
(x + 2)(x + 1)  x2 + 3x + 2 with a = 1, b = 3, and c = 2.
Hence, "to factor a trinomial" means to write the trinomial as a
product of two binomials, that is, to convert
ax2 + bx + c  (#x + #)(#x + #)
Factoring Trinomials and Making Lists

6. For our discussions, trinomials (three-term) in x are polynomials
of the form ax2 + bx + c where a (≠ 0), b, and c are numbers.
In general, we have (#x + #)(#x + #)  ax2 + bx + c.
For example,
(x + 2)(x + 1)  x2 + 3x + 2 with a = 1, b = 3, and c = 2.
Hence, "to factor a trinomial" means to write the trinomial as a
product of two binomials, that is, to convert
ax2 + bx + c  (#x + #)(#x + #)
Factoring Trinomials and Making Lists
The Basic Fact About Factoring Trinomials:

7. For our discussions, trinomials (three-term) in x are polynomials
of the form ax2 + bx + c where a (≠ 0), b, and c are numbers.
In general, we have (#x + #)(#x + #)  ax2 + bx + c.
For example,
(x + 2)(x + 1)  x2 + 3x + 2 with a = 1, b = 3, and c = 2.
Hence, "to factor a trinomial" means to write the trinomial as a
product of two binomials, that is, to convert
ax2 + bx + c  (#x + #)(#x + #)
Factoring Trinomials and Making Lists
The Basic Fact About Factoring Trinomials:
There are two types of trinomials,
l. the ones that are factorable such as
x2 + 3x + 2  (x + 2)(x + 1)

8. For our discussions, trinomials (three-term) in x are polynomials
of the form ax2 + bx + c where a (≠ 0), b, and c are numbers.
In general, we have (#x + #)(#x + #)  ax2 + bx + c.
For example,
(x + 2)(x + 1)  x2 + 3x + 2 with a = 1, b = 3, and c = 2.
Hence, "to factor a trinomial" means to write the trinomial as a
product of two binomials, that is, to convert
ax2 + bx + c  (#x + #)(#x + #)
Factoring Trinomials and Making Lists
The Basic Fact About Factoring Trinomials:
There are two types of trinomials,
l. the ones that are factorable such as
x2 + 3x + 2  (x + 2)(x + 1)
ll. the ones that are prime or no factorable, such as
x2 + 2x + 3  (#x + #)(#x + #) (Not possible!)

9. For our discussions, trinomials (three-term) in x are polynomials
of the form ax2 + bx + c where a (≠ 0), b, and c are numbers.
In general, we have (#x + #)(#x + #)  ax2 + bx + c.
For example,
(x + 2)(x + 1)  x2 + 3x + 2 with a = 1, b = 3, and c = 2.
Hence, "to factor a trinomial" means to write the trinomial as a
product of two binomials, that is, to convert
ax2 + bx + c  (#x + #)(#x + #)
Factoring Trinomials and Making Lists
The Basic Fact About Factoring Trinomials:
There are two types of trinomials,
l. the ones that are factorable such as
x2 + 3x + 2  (x + 2)(x + 1)
ll. the ones that are prime or no factorable, such as
x2 + 2x + 3  (#x + #)(#x + #)
Our jobs are to determine which trinomials:
1. are factorable and factor them,
(Not possible!)

10. For our discussions, trinomials (three-term) in x are polynomials
of the form ax2 + bx + c where a (≠ 0), b, and c are numbers.
In general, we have (#x + #)(#x + #)  ax2 + bx + c.
For example,
(x + 2)(x + 1)  x2 + 3x + 2 with a = 1, b = 3, and c = 2.
Hence, "to factor a trinomial" means to write the trinomial as a
product of two binomials, that is, to convert
ax2 + bx + c  (#x + #)(#x + #)
Factoring Trinomials and Making Lists
The Basic Fact About Factoring Trinomials:
There are two types of trinomials,
l. the ones that are factorable such as
x2 + 3x + 2  (x + 2)(x + 1)
ll. the ones that are prime or no factorable, such as
x2 + 2x + 3  (#x + #)(#x + #)
Our jobs are to determine which trinomials:
1. are factorable and factor them,
2. are prime so we won’t waste time on trying to factor them.
(Not possible!)

11. Factoring Trinomials and Making Lists
One method to determine which is which is by making lists.

12. Factoring Trinomials and Making Lists
One method to determine which is which is by making lists.
A list is a record of all the possibilities according to some criteria
such as the list of “all the cousins that one has”.

13. Factoring Trinomials and Making Lists
One method to determine which is which is by making lists.
A list is a record of all the possibilities according to some criteria
such as the list of “all the cousins that one has”.
The lists we will make are lists of numbers.

14. Example A.
Using the given tables,
list all the u and v such that:
Factoring Trinomials and Making Lists
One method to determine which is which is by making lists.
A list is a record of all the possibilities according to some criteria
such as the list of “all the cousins that one has”.
The lists we will make are lists of numbers.
12 12
I
i. uv is the top number
II

15. Example A.
Using the given tables,
list all the u and v such that:
Factoring Trinomials and Making Lists
One method to determine which is which is by making lists.
A list is a record of all the possibilities according to some criteria
such as the list of “all the cousins that one has”.
The lists we will make are lists of numbers.
12
7
12
9
I
ii. and if possible,
u + v is the bottom number.
i. uv is the top number
II

16. Example A.
Using the given tables,
list all the u and v such that:
Factoring Trinomials and Making Lists
One method to determine which is which is by making lists.
A list is a record of all the possibilities according to some criteria
such as the list of “all the cousins that one has”.
The lists we will make are lists of numbers.
12
7
We list all the possible ways
to factor 12 as u*v as shown. 12
9
I
1 12
6
3 4
2
1 12
6
3 4
2
ii. and if possible,
u + v is the bottom number.
i. uv is the top number
II

17. Example A.
Using the given tables,
list all the u and v such that:
Factoring Trinomials and Making Lists
One method to determine which is which is by making lists.
A list is a record of all the possibilities according to some criteria
such as the list of “all the cousins that one has”.
The lists we will make are lists of numbers.
12
7
We list all the possible ways
to factor 12 as u*v as shown.
For l, the solution are 3 and 4.
12
9
I
1 12
6
3 4
2
1 12
6
3 4
2
ii. and if possible,
u + v is the bottom number.
i. uv is the top number
II

18. Example A.
Using the given tables,
list all the u and v such that:
Factoring Trinomials and Making Lists
One method to determine which is which is by making lists.
A list is a record of all the possibilities according to some criteria
such as the list of “all the cousins that one has”.
The lists we will make are lists of numbers.
12
7
We list all the possible ways
to factor 12 as u*v as shown.
For l, the solution are 3 and 4.
For ll, based on the list of all
the possible u and v,
there are no u and v
where u + v = 9,
so the task is impossible.
12
9
I
1 12
6
3 4
2
1 12
6
3 4
2
ii. and if possible,
u + v is the bottom number.
i. uv is the top number
II

19. The ac-Method
A table like the ones above can be made from a given trinomial
and the ac–method uses the table to check if the given
trinomial is factorable or prime.
I. If we find the u and v that fit the table then it is factorable,
and we may use the grouping method, with the found u and v,
to factor the trinomial.
Factoring Trinomials and Making Lists

20. The ac-Method
A table like the ones above can be made from a given trinomial
and the ac–method uses the table to check if the given
trinomial is factorable or prime.
I. If we find the u and v that fit the table then it is factorable,
and we may use the grouping method, with the found u and v,
to factor the trinomial.
Example B. Factor x2 – x – 6 by grouping.
Here is an example of factoring a trinomial by grouping.
Factoring Trinomials and Making Lists

21. The ac-Method
A table like the ones above can be made from a given trinomial
and the ac–method uses the table to check if the given
trinomial is factorable or prime.
I. If we find the u and v that fit the table then it is factorable,
and we may use the grouping method, with the found u and v,
to factor the trinomial.
Example B. Factor x2 – x – 6 by grouping.
x2 – x – 6 write –x as –3x + 2x
= x2 – 3x + 2x – 6
Here is an example of factoring a trinomial by grouping.
Factoring Trinomials and Making Lists

22. The ac-Method
A table like the ones above can be made from a given trinomial
and the ac–method uses the table to check if the given
trinomial is factorable or prime.
I. If we find the u and v that fit the table then it is factorable,
and we may use the grouping method, with the found u and v,
to factor the trinomial.
Example B. Factor x2 – x – 6 by grouping.
x2 – x – 6 write –x as –3x + 2x
= x2 – 3x + 2x – 6 put the four terms into two pairs
= (x2 – 3x) + (2x – 6)
Here is an example of factoring a trinomial by grouping.
Factoring Trinomials and Making Lists

23. The ac-Method
A table like the ones above can be made from a given trinomial
and the ac–method uses the table to check if the given
trinomial is factorable or prime.
I. If we find the u and v that fit the table then it is factorable,
and we may use the grouping method, with the found u and v,
to factor the trinomial.
Example B. Factor x2 – x – 6 by grouping.
x2 – x – 6 write –x as –3x + 2x
= x2 – 3x + 2x – 6 put the four terms into two pairs
= (x2 – 3x) + (2x – 6) take out the GCF of each pair
= x(x – 3) + 2(x – 3)
Here is an example of factoring a trinomial by grouping.
Factoring Trinomials and Making Lists

24. The ac-Method
A table like the ones above can be made from a given trinomial
and the ac–method uses the table to check if the given
trinomial is factorable or prime.
I. If we find the u and v that fit the table then it is factorable,
and we may use the grouping method, with the found u and v,
to factor the trinomial.
Example B. Factor x2 – x – 6 by grouping.
x2 – x – 6 write –x as –3x + 2x
= x2 – 3x + 2x – 6 put the four terms into two pairs
= (x2 – 3x) + (2x – 6) take out the GCF of each pair
= x(x – 3) + 2(x – 3) take out the common (x – 3)
= (x – 3)(x + 2)
Here is an example of factoring a trinomial by grouping.
Factoring Trinomials and Making Lists

25. The ac-Method
A table like the ones above can be made from a given trinomial
and the ac–method uses the table to check if the given
trinomial is factorable or prime.
I. If we find the u and v that fit the table then it is factorable,
and we may use the grouping method, with the found u and v,
to factor the trinomial.
II. If the table is impossible to do, then the trinomial is prime.
Example B. Factor x2 – x – 6 by grouping.
x2 – x – 6 write –x as –3x + 2x
= x2 – 3x + 2x – 6 put the four terms into two pairs
= (x2 – 3x) + (2x – 6) take out the GCF of each pair
= x(x – 3) + 2(x – 3) take out the common (x – 3)
= (x – 3)(x + 2)
Here is an example of factoring a trinomial by grouping.
Factoring Trinomials and Making Lists

26. The ac-Method
A table like the ones above can be made from a given trinomial
and the ac–method uses the table to check if the given
trinomial is factorable or prime.
I. If we find the u and v that fit the table then it is factorable,
and we may use the grouping method, with the found u and v,
to factor the trinomial.
II. If the table is impossible to do, then the trinomial is prime.
Example B. Factor x2 – x – 6 by grouping.
x2 – x – 6 write –x as –3x + 2x
= x2 – 3x + 2x – 6 put the four terms into two pairs
= (x2 – 3x) + (2x – 6) take out the GCF of each pair
= x(x – 3) + 2(x – 3) take out the common (x – 3)
= (x – 3)(x + 2)
Here is an example of factoring a trinomial by grouping.
Let’s see how the X–table is made from a trinomial.
Factoring Trinomials and Making Lists

27. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is: ac
b
Factoring Trinomials and Making Lists

28. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is:
i.e. ac at the top, and b at the bottom,
ac
b
Factoring Trinomials and Making Lists

29. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is:
i.e. ac at the top, and b at the bottom,
In example B, the ac-table for 1x2 – x – 6 is:
ac
b
Factoring Trinomials and Making Lists
–6
–1

30. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is:
i.e. ac at the top, and b at the bottom,
and we are to find u and v such that
uv = ac
u + v = b
In example B, the ac-table for 1x2 – x – 6 is:
ac
b
Factoring Trinomials and Making Lists
–6
–1
u v

31. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is:
i.e. ac at the top, and b at the bottom,
and we are to find u and v such that
–6
–1
–3 2
uv = ac
u + v = b
In example B, the ac-table for 1x2 – x – 6 is:
We found –3, 2 fit the table,
ac
b
Factoring Trinomials and Making Lists
u v

32. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is:
i.e. ac at the top, and b at the bottom,
and we are to find u and v such that
–6
–1
–3 2
uv = ac
u + v = b
I. If u and v are found (so u + v = b),
write ax2 + bx + c as ax2 + ux + vx + c,
In example B, the ac-table for 1x2 – x – 6 is:
We found –3, 2 fit the table,
ac
b
Factoring Trinomials and Making Lists
u v

33. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is:
i.e. ac at the top, and b at the bottom,
and we are to find u and v such that
–6
–1
–3 2
uv = ac
u + v = b
I. If u and v are found (so u + v = b),
write ax2 + bx + c as ax2 + ux + vx + c,
In example B, the ac-table for 1x2 – x – 6 is:
We found –3, 2 fit the table, so we write
x2 – x – 6 as x2 – 3x + 2x – 6
ac
b
Factoring Trinomials and Making Lists
u v

34. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is:
i.e. ac at the top, and b at the bottom,
and we are to find u and v such that
–6
–1
–3 2
uv = ac
u + v = b
I. If u and v are found (so u + v = b),
write ax2 + bx + c as ax2 + ux + vx + c,
then factor (ax2 + ux) + (vx + c) by the grouping method.
In example B, the ac-table for 1x2 – x – 6 is:
We found –3, 2 fit the table, so we write
x2 – x – 6 as x2 – 3x + 2x – 6
ac
b
Factoring Trinomials and Making Lists
u v

35. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is:
i.e. ac at the top, and b at the bottom,
and we are to find u and v such that
–6
–1
–3 2
uv = ac
u + v = b
I. If u and v are found (so u + v = b),
write ax2 + bx + c as ax2 + ux + vx + c,
then factor (ax2 + ux) + (vx + c) by the grouping method.
In example B, the ac-table for 1x2 – x – 6 is:
We found –3, 2 fit the table, so we write
x2 – x – 6 as x2 – 3x + 2x – 6 and by grouping
= (x2 – 3x) + (2x – 6)
ac
b
Factoring Trinomials and Making Lists
u v

36. ac-Method: Given the trinomial ax2 + bx + c
with no common factor, it’s ac–table is:
i.e. ac at the top, and b at the bottom,
and we are to find u and v such that
–6
–1
–3 2
uv = ac
u + v = b
I. If u and v are found (so u + v = b),
write ax2 + bx + c as ax2 + ux + vx + c,
then factor (ax2 + ux) + (vx + c) by the grouping method.
In example B, the ac-table for 1x2 – x – 6 is:
We found –3, 2 fit the table, so we write
x2 – x – 6 as x2 – 3x + 2x – 6 and by grouping
= (x2 – 3x) + (2x – 6)
= x(x – 3) + 2(x – 3)
= (x – 3)(x + 2)
ac
b
u v
Factoring Trinomials and Making Lists

37. Example C. Factor 3x2 – 4x – 20 using the ac-method.
Factoring Trinomials and Making Lists

38. Example C. Factor 3x2 – 4x – 20 using the ac-method.
We have that a = 3, c = –20 so ac = 3(–20) = –60,
Factoring Trinomials and Making Lists

39. Example C. Factor 3x2 – 4x – 20 using the ac-method.
We have that a = 3, c = –20 so ac = 3(–20) = –60,
b = –4 and the ac–table is:
–60
–4
Factoring Trinomials and Making Lists

40. Example C. Factor 3x2 – 4x – 20 using the ac-method.
We have that a = 3, c = –20 so ac = 3(–20) = –60,
b = –4 and the ac–table is:
We need two numbers u and v such that
uv = –60 and u + v = –4.
–60
–4
Factoring Trinomials and Making Lists

41. Example C. Factor 3x2 – 4x – 20 using the ac-method.
We have that a = 3, c = –20 so ac = 3(–20) = –60,
b = –4 and the ac–table is:
We need two numbers u and v such that
uv = –60 and u + v = –4.
By trial and error we see that 6 and –10 is the
solution so we may factor the trinomial by grouping.
–60
–4
–10 6
Factoring Trinomials and Making Lists

42. Example C. Factor 3x2 – 4x – 20 using the ac-method.
We have that a = 3, c = –20 so ac = 3(–20) = –60,
b = –4 and the ac–table is:
We need two numbers u and v such that
uv = –60 and u + v = –4.
By trial and error we see that 6 and –10 is the
solution so we may factor the trinomial by grouping.
–60
–4
–10 6
Factoring Trinomials and Making Lists
Using 6 and –10, write 3x2 – 4x – 20 as 3x2 + 6x –10x – 20

43. Example C. Factor 3x2 – 4x – 20 using the ac-method.
We have that a = 3, c = –20 so ac = 3(–20) = –60,
b = –4 and the ac–table is:
We need two numbers u and v such that
uv = –60 and u + v = –4.
By trial and error we see that 6 and –10 is the
solution so we may factor the trinomial by grouping.
–60
–4
–10 6
Factoring Trinomials and Making Lists
Using 6 and –10, write 3x2 – 4x – 20 as 3x2 + 6x –10x – 20
= (3x2 + 6x ) + (–10x – 20) put in two groups

44. Example C. Factor 3x2 – 4x – 20 using the ac-method.
We have that a = 3, c = –20 so ac = 3(–20) = –60,
b = –4 and the ac–table is:
We need two numbers u and v such that
uv = –60 and u + v = –4.
By trial and error we see that 6 and –10 is the
solution so we may factor the trinomial by grouping.
–60
–4
–10 6
Factoring Trinomials and Making Lists
Using 6 and –10, write 3x2 – 4x – 20 as 3x2 + 6x –10x – 20
= (3x2 + 6x ) + (–10x – 20) put in two groups
= 3x(x + 2) – 10 (x + 2) pull out common factor

45. Example C. Factor 3x2 – 4x – 20 using the ac-method.
We have that a = 3, c = –20 so ac = 3(–20) = –60,
b = –4 and the ac–table is:
We need two numbers u and v such that
uv = –60 and u + v = –4.
By trial and error we see that 6 and –10 is the
solution so we may factor the trinomial by grouping.
–60
–4
–10 6
Factoring Trinomials and Making Lists
Using 6 and –10, write 3x2 – 4x – 20 as 3x2 + 6x –10x – 20
= (3x2 + 6x ) + (–10x – 20) put in two groups
= 3x(x + 2) – 10 (x + 2) pull out common factor
= (3x – 10)(x + 2) pull out common factor
Hence 3x2 – 4x – 20 = (3x – 10)(x + 2)

46. Example C. Factor 3x2 – 4x – 20 using the ac-method.
If the trinomial is prime then we have to justify it’s prime by
showing that no such u and v exist
We have that a = 3, c = –20 so ac = 3(–20) = –60,
b = –4 and the ac–table is:
We need two numbers u and v such that
uv = –60 and u + v = –4.
By trial and error we see that 6 and –10 is the
solution so we may factor the trinomial by grouping.
–60
–4
–10 6
Factoring Trinomials and Making Lists
Using 6 and –10, write 3x2 – 4x – 20 as 3x2 + 6x –10x – 20
= (3x2 + 6x ) + (–10x – 20) put in two groups
= 3x(x + 2) – 10 (x + 2) pull out common factor
= (3x – 10)(x + 2) pull out common factor
Hence 3x2 – 4x – 20 = (3x – 10)(x + 2)

47. Example C. Factor 3x2 – 4x – 20 using the ac-method.
If the trinomial is prime then we have to justify it’s prime by
showing that no such u and v exist by listing all the possible
u’s and v’s such that uv = ac in the table to demonstrate that
none of them fits the condition u + v = b.
We have that a = 3, c = –20 so ac = 3(–20) = –60,
b = –4 and the ac–table is:
We need two numbers u and v such that
uv = –60 and u + v = –4.
By trial and error we see that 6 and –10 is the
solution so we may factor the trinomial by grouping.
–60
–4
–10 6
Factoring Trinomials and Making Lists
Using 6 and –10, write 3x2 – 4x – 20 as 3x2 + 6x –10x – 20
= (3x2 + 6x ) + (–10x – 20) put in two groups
= 3x(x + 2) – 10 (x + 2) pull out common factor
= (3x – 10)(x + 2) pull out common factor
Hence 3x2 – 4x – 20 = (3x – 10)(x + 2)

48. Example D. Factor 3x2 – 6x – 20 if possible.
If it’s prime, justify that.
Factoring Trinomials and Making Lists

49. Example D. Factor 3x2 – 6x – 20 if possible.
If it’s prime, justify that.
a = 3, c = –20, hence ac = 3(–20) = –60,
with b = –6, we have the ac–table:
Factoring Trinomials and Making Lists
–60
–6

50. Example D. Factor 3x2 – 6x – 20 if possible.
If it’s prime, justify that.
a = 3, c = –20, hence ac = 3(–20) = –60,
with b = –6, we have the ac–table:
We want two numbers u and v such that
uv = –60 and u + v = –6.
Factoring Trinomials and Making Lists
–60
–6

51. Example D. Factor 3x2 – 6x – 20 if possible.
If it’s prime, justify that.
a = 3, c = –20, hence ac = 3(–20) = –60,
with b = –6, we have the ac–table:
We want two numbers u and v such that
uv = –60 and u + v = –6.
After failing to guess two such numbers,
we check to see if it's prime by listing in order
all positive u’s and v’s where uv = 60 as shown.
Factoring Trinomials and Making Lists
–60
–6

52. Example D. Factor 3x2 – 6x – 20 if possible.
If it’s prime, justify that.
a = 3, c = –20, hence ac = 3(–20) = –60,
with b = –6, we have the ac–table:
We want two numbers u and v such that
uv = –60 and u + v = –6.
After failing to guess two such numbers,
we check to see if it's prime by listing in order
all positive u’s and v’s where uv = 60 as shown.
Factoring Trinomials and Making Lists
–60
–6
601
302
203
154
125
106
Always make a
list in an orderly
manner to ensure
the accuracy of
the list.

53. Example D. Factor 3x2 – 6x – 20 if possible.
If it’s prime, justify that.
a = 3, c = –20, hence ac = 3(–20) = –60,
with b = –6, we have the ac–table:
We want two numbers u and v such that
uv = –60 and u + v = –6.
After failing to guess two such numbers,
we check to see if it's prime by listing in order
all positive u’s and v’s where uv = 60 as shown.
By the table, we see that there are no u and v
such that (±) u and v combine to be –6.
Factoring Trinomials and Making Lists
–60
–6
601
302
203
154
125
106
Always make a
list in an orderly
manner to ensure
the accuracy of
the list.

54. Example D. Factor 3x2 – 6x – 20 if possible.
If it’s prime, justify that.
a = 3, c = –20, hence ac = 3(–20) = –60,
with b = –6, we have the ac–table:
We want two numbers u and v such that
uv = –60 and u + v = –6.
After failing to guess two such numbers,
we check to see if it's prime by listing in order
all positive u’s and v’s where uv = 60 as shown.
By the table, we see that there are no u and v
such that (±) u and v combine to be –6.
Hence 3x2 – 6x – 20 is prime.
Factoring Trinomials and Making Lists
–60
–6
601
302
203
154
125
106
Always make a
list in an orderly
manner to ensure
the accuracy of
the list.

55. Example D. Factor 3x2 – 6x – 20 if possible.
If it’s prime, justify that.
a = 3, c = –20, hence ac = 3(–20) = –60,
with b = –6, we have the ac–table:
We want two numbers u and v such that
uv = –60 and u + v = –6.
After failing to guess two such numbers,
we check to see if it's prime by listing in order
all positive u’s and v’s where uv = 60 as shown.
By the table, we see that there are no u and v
such that (±) u and v combine to be –6.
Hence 3x2 – 6x – 20 is prime.
Factoring Trinomials and Making Lists
–60
–6
601
302
203
154
125
106
Always make a
list in an orderly
manner to ensure
the accuracy of
the list.
Factoring By Trial and Error

56. Example D. Factor 3x2 – 6x – 20 if possible.
If it’s prime, justify that.
a = 3, c = –20, hence ac = 3(–20) = –60,
with b = –6, we have the ac–table:
We want two numbers u and v such that
uv = –60 and u + v = –6.
After failing to guess two such numbers,
we check to see if it's prime by listing in order
all positive u’s and v’s where uv = 60 as shown.
By the table, we see that there are no u and v
such that (±) u and v combine to be –6.
Hence 3x2 – 6x – 20 is prime.
Factoring Trinomials and Making Lists
–60
–6
601
302
203
154
125
106
Always make a
list in an orderly
manner to ensure
the accuracy of
the list.
Finally for some trinomials, such as when a = 1 or x2 + bx + c,
it’s easier to guess directly because it must factor into the form
(x ± u) (x ± v) if it’s factorable.
Factoring By Trial and Error

57. Example E.
a. Factor x2 + 5x + 6
By Trial and Error

58. Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6
By Trial and Error

59. Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
By Trial and Error

60. Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3)
By Trial and Error

61. Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x
By Trial and Error

62. Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
By Trial and Error

63. Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

64. b. Factor x2 – 5x + 6
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

65. b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6,
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

66. b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

67. b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

68. b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3)
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

69. b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

70. b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

71. c. Factor x2 + 5x – 6
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

72. c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

73. c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and
u + v = 5.
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

74. c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and
u + v = 5.
Since -6 = (–1)(6) = (1)(–6) = (–2)(3) =(2)(–3)
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

75. c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and
u + v = 5.
Since -6 = (–1)(6) = (1)(–6) = (–2)(3) =(2)(–3) and –1 + 6 = 5,
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

76. c. Factor x2 + 5x – 6
We want (x + u)(x + v) = x2 + 5x – 6, so we need uv = –6 and
u + v = 5.
Since -6 = (–1)(6) = (1)(–6) = (–2)(3) =(2)(–3) and –1 + 6 = 5,
so x2 + 5x – 6 = (x – 1)(x + 6).
b. Factor x2 – 5x + 6
We want (x + u)(x + v) = x2 – 5x + 6, so we need u and v
where uv = 6 and u + v = –5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and –2 – 3 = –5,
so x2 – 5x + 6 = (x – 2)(x – 3).
Example E.
a. Factor x2 + 5x + 6
We want (x + u)(x + v) = x2 + 5x + 6, so we need u and v
where uv = 6 and u + v = 5.
Since 6 = (1)(6) = (2)(3) = (-1)(-6) = (-2)(-3) and 2x + 3x = 5x,
so x2 + 5x + 6 = (x + 2)(x + 3)
By Trial and Error

77. This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.
By Trial and Error

78. Example F. Factor 3x2 + 5x + 2.
By Trial and Error
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

79. Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
By Trial and Error
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

80. Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2
By Trial and Error
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

81. Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2 to get the constant term +2.
By Trial and Error
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

82. Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
By Trial and Error
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

83. Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x ± #)(1x ± #) must yields +5x,
By Trial and Error
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

84. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
By Trial and Error
Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x ± #)(1x ± #) must yields +5x, or that
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

85. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Since 3(1) +1(2) = 5,
By Trial and Error
Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x ± #)(1x ± #) must yields +5x, or that
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

86. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Since 3(1) +1(2) = 5,
By Trial and Error
Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x ± #)(1x ± #) must yields +5x, or that
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

87. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Since 3(1) +1(2) = 5,
By Trial and Error
Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x ± #)(1x ± #) must yields +5x, or that
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

88. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Since 3(1) +1(2) = 5, we see that
3x2 + 5x + 2 = (3x + 2)(1x + 1).
By Trial and Error
Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x ± #)(1x ± #) must yields +5x, or that
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

89. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Since 3(1) +1(2) = 5, we see that
3x2 + 5x + 2 = (3x + 2)(1x + 1).
5x
By Trial and Error
Example F. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2 to get the constant term +2.
We need to place 1 and 2 as the #'s so the product will
yield the correct middle term +5x.
That is, (3x ± #)(1x ± #) must yields +5x, or that
This reversed–FOIL method (by trial and error) is useful when
the numbers involved can only be factored in few options.

90. Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
Factoring Trinomials and Making Lists

91. Example G. Use b2 – 4ac to check if the trinomial is factorable.
a. 3x2 – 7x + 2
Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
Factoring Trinomials and Making Lists
b. 3x2 – 7x – 2

92. Example G. Use b2 – 4ac to check if the trinomial is factorable.
Theorem: The trinomial ax2 + bx + c is factorable
if b2 – 4ac = 0, 1, 4, 9, 16, 25, ..i.e. it’s a squared number.
a. 3x2 – 7x + 2
Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
Factoring Trinomials and Making Lists
b. 3x2 – 7x – 2

93. Example G. Use b2 – 4ac to check if the trinomial is factorable.
Theorem: The trinomial ax2 + bx + c is factorable
if b2 – 4ac = 0, 1, 4, 9, 16, 25, ..i.e. it’s a squared number.
a. 3x2 – 7x + 2
Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
Factoring Trinomials and Making Lists
b. 3x2 – 7x – 2
a = 3, b = (–7) and c = 2

94. Example G. Use b2 – 4ac to check if the trinomial is factorable.
Theorem: The trinomial ax2 + bx + c is factorable
if b2 – 4ac = 0, 1, 4, 9, 16, 25, ..i.e. it’s a squared number.
a. 3x2 – 7x + 2
b2 – 4ac
= (–7)2 – 4(3)(2)
Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
Factoring Trinomials and Making Lists
a = 3, b = (–7) and c = 2
b. 3x2 – 7x – 2

95. Example G. Use b2 – 4ac to check if the trinomial is factorable.
Theorem: The trinomial ax2 + bx + c is factorable
if b2 – 4ac = 0, 1, 4, 9, 16, 25, ..i.e. it’s a squared number.
a. 3x2 – 7x + 2
b2 – 4ac
= (–7)2 – 4(3)(2)
= 49 – 24
= 25 which is a squared number, hence it is factorable.
Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
Factoring Trinomials and Making Lists
b. 3x2 – 7x – 2
a = 3, b = (–7) and c = 2

96. Example G. Use b2 – 4ac to check if the trinomial is factorable.
Theorem: The trinomial ax2 + bx + c is factorable
if b2 – 4ac = 0, 1, 4, 9, 16, 25, ..i.e. it’s a squared number.
If b2 – 4ac = not a squared number, then it’s not factorable.
a. 3x2 – 7x + 2
b2 – 4ac
= (–7)2 – 4(3)(2)
= 49 – 24
= 25 which is a squared number, hence it is factorable.
Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
Factoring Trinomials and Making Lists
b. 3x2 – 7x – 2
a = 3, b = (–7) and c = 2

97. Example G. Use b2 – 4ac to check if the trinomial is factorable.
Theorem: The trinomial ax2 + bx + c is factorable
if b2 – 4ac = 0, 1, 4, 9, 16, 25, ..i.e. it’s a squared number.
If b2 – 4ac = not a squared number, then it’s not factorable.
a. 3x2 – 7x + 2
b2 – 4ac
= (–7)2 – 4(3)(2)
= 49 – 24
= 25 which is a squared number, hence it is factorable.
Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
Factoring Trinomials and Making Lists
b. 3x2 – 7x – 2
a = 3, b = (–7) and c = 2
a = 3, b = (–7) and c = (–2)

98. Example G. Use b2 – 4ac to check if the trinomial is factorable.
b2 – 4ac
= (–7)2 – 4(3)(–2)
Theorem: The trinomial ax2 + bx + c is factorable
if b2 – 4ac = 0, 1, 4, 9, 16, 25, ..i.e. it’s a squared number.
If b2 – 4ac = not a squared number, then it’s not factorable.
a. 3x2 – 7x + 2
b2 – 4ac
= (–7)2 – 4(3)(2)
= 49 – 24
= 25 which is a squared number, hence it is factorable.
Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
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b. 3x2 – 7x – 2
a = 3, b = (–7) and c = 2
a = 3, b = (–7) and c = (–2)

99. Example G. Use b2 – 4ac to check if the trinomial is factorable.
b2 – 4ac
= (–7)2 – 4(3)(–2)
= 49 + 24
= 73 is not a square, hence it is prime.
Theorem: The trinomial ax2 + bx + c is factorable
if b2 – 4ac = 0, 1, 4, 9, 16, 25, ..i.e. it’s a squared number.
If b2 – 4ac = not a squared number, then it’s not factorable.
a. 3x2 – 7x + 2
b2 – 4ac
= (–7)2 – 4(3)(2)
= 49 – 24
= 25 which is a squared number, hence it is factorable.
Besides the ac–method, here is another method that’s based
on a calculating a number to check if a trinomial is factorable.
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b. 3x2 – 7x – 2
a = 3, b = (–7) and c = 2
a = 3, b = (–7) and c = (–2)

100. Observations About Signs
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101. Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
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102. Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
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103. Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
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104. Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
Factoring Trinomials and Making Lists

105. {
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
x2 – 5x + 6 = (x – 2)(x – 3)
Factoring Trinomials and Making Lists

106. {
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
x2 – 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs.
Factoring Trinomials and Making Lists

107. {
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
x2 – 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs. The
one with larger absolute value has the same sign as b.
Factoring Trinomials and Making Lists

108. {
Observations About Signs
Given that x2 + bx + c = (x + u)(x + v) so that uv = c,
we observe the following.
1. If c is positive, then u and v have same sign.
In particular,
if b is also positive, then both are positive.
if b is negative, then both are negative.
From the examples above
x2 + 5x + 6 = (x + 2)(x + 3)
x2 – 5x + 6 = (x – 2)(x – 3)
2. If c is negative, then u and v have opposite signs. The
one with larger absolute value has the same sign as b.
From the example above
x2 – 5x – 6 = (x – 6)(x + 1)
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109. 1. 3x2 – x – 2 2. 3x2 + x – 2 3. 3x2 – 2x – 1
4. 3x2 + 2x – 1 5. 2x2 – 3x + 1 6. 2x2 + 3x – 1
8. 2x2 – 3x – 27. 2x2 + 3x – 2
15. 6x2 + 5x – 6
10. 5x2 + 9x – 2
B. Factor. Factor out the GCF, the “–”, and arrange the
terms in order first.
9. 5x2 – 3x – 2
12. 3x2 – 5x + 211. 3x2 + 5x + 2
14. 6x2 – 5x – 613. 3x2 – 5x + 2
16. 6x2 – x – 2 17. 6x2 – 13x + 2 18. 6x2 – 13x + 2
19. 6x2 + 7x + 2 20. 6x2 – 7x + 2 21. 6x2 – 13x + 6
22. 6x2 + 13x + 6 23. 6x2 – 5x – 4 24. 6x2 – 13x + 8
25. 6x2 – 13x – 8 25. 4x2 – 9 26. 4x2 – 49
27. 25x2 – 4 28. 4x2 + 9 29. 25x2 + 9
30. – 6x2 – 5xy + 6y2 31. – 3x2 + 2x3– 2x 32. –6x3 – x2 + 2x
33. –15x3 – 25x2 – 10x 34. 12x3y2 –14x2y2 + 4xy2
Exercise A. Use the ac–method, factor the trinomial or
demonstrate that it’s not factorable.
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110. 35. –3x3 – 30x2 – 48x34. –yx2 + 4yx + 5y
36. –2x3 + 20x2 – 24x
40. 4x2 – 44xy + 96y2
37. –x2 + 11xy + 24y2
38. x4 – 6x3 + 36x2 39. –x2 + 9xy + 36y2
C. Factor. Factor out the GCF, the “–”, and arrange the
terms in order first.
D. Factor. If not possible, state so.
41. x2 + 1 42. x2 + 4 43. x2 + 9 43. 4x2 + 25
44. What can you conclude from 41–43?
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