The document discusses factoring trinomials of the form ax^2 + bx + c by using the reversed FOIL method. It explains that this involves finding four numbers that when multiplied using FOIL will yield the trinomial. Examples are provided to show how to factor trinomials by setting up equations from the FOIL factors and constants and solving to find the numbers. The goal is to find the numbers that satisfy the equations to factor the trinomial.

1. Factoring Trinomials III

2. Factoring Trinomials III
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

3. Factoring Trinomials III
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

4. Factoring Trinomials III
For this method, we need to find four numbers that fit certain
descriptions.
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

5. Factoring Trinomials III
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

6. Factoring Trinomials III
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

7. Factoring Trinomials III
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (± ) + 3*(± ) = 5.
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

8. Factoring Trinomials III
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (± ) + 3*(± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

9. Factoring Trinomials III
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (± ) + 3*(± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (± ) + 3* (± ) = –5.
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

10. Factoring Trinomials III
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (± ) + 3*(± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (± ) + 3* (± ) = –5.
Yes, 1* (1) + 3* (–2) = –5
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

11. Factoring Trinomials III
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (± ) + 3*(± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (± ) + 3* (± ) = –5.
Yes, 1* (1) + 3* (–2) = –5 or 1* (–2) + 3* (–1) = –5
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

12. Factoring Trinomials III
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (± ) + 3*(± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (± ) + 3* (± ) = –5.
Yes, 1* (1) + 3* (–2) = –5 or 1* (–2) + 3* (–1) = –5
c. 1* (± ) + 3* (± ) = 8.
Reversed FOIL Method
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.

13. Factoring Trinomials III
Too factor trinomials of the form ax2 + bx + c, the shortest,
but not a reliable way, is by guessing, .i.e. we reverse the
FOIL procedure and try to guess what numbers are needed.
For this method, we need to find four numbers that fit certain
descriptions. The following are examples of the task to be
accomplished.
Example A. Let {1, 3} and {1, 2} be two pairs of numbers.
Is it possible to split the {1, 2 }, put them in the boxes that
makes the equality true?
a. 1* (± ) + 3*(± ) = 5.
Yes, 1* (2) + 3 * (1) = 5
b. 1* (± ) + 3* (± ) = –5.
Yes, 1* (1) + 3* (–2) = –5 or 1* (–2) + 3* (–1) = –5
c. 1* (± ) + 3* (± ) = 8.
No, since the most we can obtain is 1* (1) + 3* (2) = 7.
Reversed FOIL Method

14. Factoring Trinomials III
(Reversed FOIL Method)

15. Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

16. Example B. Factor 3x2 + 5x + 2.
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

17. Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

18. Example B. Factor 3x2 + 5x + 2.
The only way to get 3x2 is (3x ± #)(1x ± #).
The #’s must be 1 and 2
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

19. Example B. 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.
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

20. Example B. 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.
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

21. Example B. 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,
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

22. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.
Example B. 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

23. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Since 3(1) +1(2) = 5,
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.
Example B. 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

24. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Since 3(1) +1(2) = 5,
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.
Example B. 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

25. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Since 3(1) +1(2) = 5,
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.
Example B. 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

26. 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).
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.
Example B. 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

27. 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
Factoring Trinomials III
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.
Example B. 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

28. Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.

29. Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).

30. Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s

31. 3(± # ) + 1(± # ) = –7.
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

32. 3(± # ) + 1(± # ) = –7.
It's 3(–2) + 1(–1) = –7.
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

33. 3(± # ) + 1(± # ) = –7.
It's 3(–2) + 1(–1) = –7.
So 3x2 – 7x + 2 = (3x –1)(1x – 2)
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

34. 3(± # ) + 1(± # ) = –7.
It's 3(–2) + 1(–1) = –7.
So 3x2 – 7x + 2 = (3x –1)(1x – 2)
Example D. Factor 3x2 + 5x – 2.
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

35. 3(± # ) + 1(± # ) = –7.
It's 3(–2) + 1(–1) = –7.
So 3x2 – 7x + 2 = (3x –1)(1x – 2)
Example D. Factor 3x2 + 5x – 2.
We start with (3x ± #)(1x ± #).
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

36. 3(± # ) + 1(± # ) = –7.
It's 3(–2) + 1(–1) = –7.
So 3x2 – 7x + 2 = (3x –1)(1x – 2)
Example D. Factor 3x2 + 5x – 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

37. 3(± # ) + 1(± # ) = –7.
It's 3(–2) + 1(–1) = –7.
So 3x2 – 7x + 2 = (3x –1)(1x – 2)
Example D. Factor 3x2 + 5x – 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that
3(± # ) + 1(± # ) = +5.
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

38. 3(± # ) + 1(± # ) = –7.
It's 3(–2) + 1(–1) = –7.
So 3x2 – 7x + 2 = (3x –1)(1x – 2)
Example D. Factor 3x2 + 5x – 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that
3(± # ) + 1(± # ) = +5.
Since c is negative, they must have opposite signs .
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

39. 3(± # ) + 1(± # ) = –7.
It's 3(–2) + 1(–1) = –7.
So 3x2 – 7x + 2 = (3x –1)(1x – 2)
Example D. Factor 3x2 + 5x – 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that
3(± # ) + 1(± # ) = +5.
It is 3(+2) + 1(–1) = +5.
Since c is negative, they must have opposite signs .
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

40. 3(± # ) + 1(± # ) = –7.
It's 3(–2) + 1(–1) = –7.
So 3x2 – 7x + 2 = (3x –1)(1x – 2)
Example D. Factor 3x2 + 5x – 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that
3(± # ) + 1(± # ) = +5.
It is 3(+2) + 1(–1) = +5.
So 3x2 + 5x + 2 = (3x –1)(1x + 2)
Since c is negative, they must have opposite signs .
Factoring Trinomials III
Example C. Factor 3x2 – 7x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1 and 2 as #'s so that

41. Example E. Factor 3x2 + 8x + 2.
Factoring Trinomials III

42. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
Factoring Trinomials III

43. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
Factoring Trinomials III

44. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible.
Factoring Trinomials III

45. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
Factoring Trinomials III

46. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
Factoring Trinomials III
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.

47. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
Factoring Trinomials III
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x – 4.

48. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
Factoring Trinomials III
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x – 4.
We start with (3x ± #)(1x ± #).

49. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
Factoring Trinomials III
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x – 4.
We start with (3x ± #)(1x ± #). Since 4 = 2(2) = 1(4),

50. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
3(± # ) + 1(± # ) = +11.
Factoring Trinomials III
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x – 4.
We start with (3x ± #)(1x ± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that

51. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
3(± # ) + 1(± # ) = +11. It can't be 2&2.
Factoring Trinomials III
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x – 4.
We start with (3x ± #)(1x ± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that
3(± # ) + 1(± # ) = +11.

52. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
3(± # ) + 1(± # ) = +11. It can't be 2&2.
Factoring Trinomials III
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x – 4.
We start with (3x ± #)(1x ± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that
Try 1&4,
3(± # ) + 1(± # ) = +11.

53. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
3(± # ) + 1(± # ) = +11. It can't be 2&2.
Factoring Trinomials III
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x – 4.
We start with (3x ± #)(1x ± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that
Try 1&4, it is
3(+4) + 1(–1) = +11.
3(± # ) + 1(± # ) = +11.

54. Example E. Factor 3x2 + 8x + 2.
We start with (3x ± #)(1x ± #).
We need to fill in 1&2 so that
3(± # ) + 1(± # ) = +8.
This is impossible. Hence the expression is prime.
3(± # ) + 1(± # ) = +11. It can't be 2&2.
Factoring Trinomials III
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.
Example F. Factor 3x2 + 11x – 4.
We start with (3x ± #)(1x ± #). Since 4 = 2(2) = 1(4),
we need to fill in 2&2 or 1&4 as #'s so that
Try 1&4, it is
3(+4) + 1(–1) = +11.
So 3x2 + 11x – 4 = (3x – 1)(1x + 4).

55. Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.

56. Example G. Factor 12x2 – 5x – 3.
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.

57. Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1),
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.

58. Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.

59. Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
(± #)(± 3) + (± #)(±1) = – 5.

60. Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
(± #)(± 3) + (± #)(±1) = – 5.
12 = 1(12) = 2(6) = 3(4)

61. Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
(± #)(± 3) + (± #)(±1) = – 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.

62. Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
(± #)(± 3) + (± #)(±1) = – 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
We get (3)(–3) + (4)(+1) = – 5.

63. Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
So 12x2 – 5x – 3 = (3x + 1)(4x – 3).
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
(± #)(± 3) + (± #)(±1) = – 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
We get (3)(–3) + (4)(+1) = – 5.

64. Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
So 12x2 – 5x – 3 = (3x + 1)(4x – 3).
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
(± #)(± 3) + (± #)(±1) = – 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
We get (3)(–3) + (4)(+1) = – 5.
Remark:
In the above method, finding
(#)(± #) + (#)( ± #) = b
does not guarantee that the trinomial will factor.

65. Example G. Factor 12x2 – 5x – 3.
Since 3 must be 3(1), we need to find two numbers such
that (#)(#) = 12 and that
So 12x2 – 5x – 3 = (3x + 1)(4x – 3).
Factoring Trinomials III
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
(± #)(± 3) + (± #)(±1) = – 5.
12 = 1(12) = 2(6) = 3(4)
1&12 and 2&6 can be quickly eliminated.
We get (3)(–3) + (4)(+1) = – 5.
Remark:
In the above method, finding
(#)(± #) + (#)( ± #) = b
does not guarantee that the trinomial will factor. We have to
match the sign of c also.

66. Example H. Factor 3x2 – 7x – 2 .
Factoring Trinomials III

67. Example H. Factor 3x2 – 7x – 2 .
We start with (3x ± #)(1x ± #).
Factoring Trinomials III

68. Example H. Factor 3x2 – 7x – 2 .
We start with (3x ± #)(1x ± #). We find that:
3(–2) + 1(–1) = –7.
Factoring Trinomials III

69. Example H. Factor 3x2 – 7x – 2 .
We start with (3x ± #)(1x ± #). We find that:
3(–2) + 1(–1) = –7.
But this won't work since (–2)(–1) = 2 = c.
Factoring Trinomials III

70. Example H. Factor 3x2 – 7x – 2 .
We start with (3x ± #)(1x ± #). We find that:
3(–2) + 1(–1) = –7.
But this won't work since (–2)(–1) = 2 = c.
In fact this trinomial is prime.
Factoring Trinomials III

71. Example H. Factor 3x2 – 7x – 2 .
We start with (3x ± #)(1x ± #). We find that:
3(–2) + 1(–1) = –7.
But this won't work since (–2)(–1) = 2 = c.
In fact this trinomial is prime.
Factoring Trinomials III
There might be multiple matchings for
(#)(± #) + (#)( ± #) = b
make sure you chose the correct one, if any.

72. Example H. Factor 3x2 – 7x – 2 .
We start with (3x ± #)(1x ± #). We find that:
3(–2) + 1(–1) = –7.
But this won't work since (–2)(–1) = 2 = c.
In fact this trinomial is prime.
Factoring Trinomials III
Example I: Factor 1x2 + 5x – 6 .
There might be multiple matchings for
(#)(± #) + (#)( ± #) = b
make sure you chose the correct one, if any.

73. Example H. Factor 3x2 – 7x – 2 .
We start with (3x ± #)(1x ± #). We find that:
3(–2) + 1(–1) = –7.
But this won't work since (–2)(–1) = 2 = c.
In fact this trinomial is prime.
Factoring Trinomials III
Example I: Factor 1x2 + 5x – 6 .
We have:
1(+3) + 1(+2) = +5
There might be multiple matchings for
(#)(± #) + (#)( ± #) = b
make sure you chose the correct one, if any.

74. Example H. Factor 3x2 – 7x – 2 .
We start with (3x ± #)(1x ± #). We find that:
3(–2) + 1(–1) = –7.
But this won't work since (–2)(–1) = 2 = c.
In fact this trinomial is prime.
Factoring Trinomials III
Example I: Factor 1x2 + 5x – 6 .
We have:
1(+3) + 1(+2) = +5
1(+6) + 1(–1) = +5
There might be multiple matchings for
(#)(± #) + (#)( ± #) = b
make sure you chose the correct one, if any.

75. Example H. Factor 3x2 – 7x – 2 .
We start with (3x ± #)(1x ± #). We find that:
3(–2) + 1(–1) = –7.
But this won't work since (–2)(–1) = 2 = c.
In fact this trinomial is prime.
Factoring Trinomials III
Example I: Factor 1x2 + 5x – 6 .
We have:
1(+3) + 1(+2) = +5
The one that works is x2 + 5x – 6 = (x + 6)(x – 1).
1(+6) + 1(–1) = +5
There might be multiple matchings for
(#)(± #) + (#)( ± #) = b
make sure you chose the correct one, if any.

76. Factoring Trinomials III
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.

77. Factoring Trinomials III
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.

78. Factoring Trinomials III
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.

79. Factoring Trinomials III
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor –x3 + 3x + 2x2

80. Factoring Trinomials III
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor –x3 + 3x + 2x2
–x3 + 3x + 2x2 Arrange the terms in order
= –x3 + 2x2 + 3x

81. Factoring Trinomials III
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor –x3 + 3x + 2x2
–x3 + 3x + 2x2 Arrange the terms in order
= –x3 + 2x2 + 3x Factor out the GCF
= – x(x2 – 2x – 3)

82. Factoring Trinomials III
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor –x3 + 3x + 2x2
–x3 + 3x + 2x2 Arrange the terms in order
= –x3 + 2x2 + 3x Factor out the GCF
= – x(x2 – 2x – 3)
= – x(x – 3)(x + 1)

83. Factoring Trinomials III
Finally, before starting the reverse-FOIL procedure
1. make sure the terms are arranged in order.
2. if there is any common factor, pull out the GCF first.
3. make sure that x2 is positive, if not, factor out the negative
sign first.
Example J. Factor –x3 + 3x + 2x2
–x3 + 3x + 2x2 Arrange the terms in order
= –x3 + 2x2 + 3x Factor out the GCF
= – x(x2 – 2x – 3)
= – x(x – 3)(x + 1)
For factoring problems, try guessing first.
If no answer is found, use the ac-method
(or the formula) to determine if it is prime or do it by grouping,
and always check your answers.

84. Ex. A. Factor the following trinomials. If it’s prime, state so.
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
Factoring Trinomials III
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. –15x2 – 25x2 – 10x 34. 12x2y2 –14x2y2 + 4xy2