The document discusses two methods for factoring trinomials of the form ax^2 + bx + c. The first method is short but not always reliable, while the second method takes more steps but always provides a definite answer. This second method, called the reversed FOIL method, involves finding four numbers that satisfy certain properties to factor the trinomial. An example is worked out step-by-step to demonstrate how to use the reversed FOIL method to factor the trinomial 3x^2 + 5x + 2.

1. Factoring Trinomials II

2. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.

3. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods.

4. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.

5. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.

6. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
Reversed FOIL Method

7. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
For this method, we need to find four numbers that fit certain
descriptions.
Reversed FOIL Method

8. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
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

9. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
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

10. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
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

11. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
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

12. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
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

13. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
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

14. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
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

15. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
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

16. Factoring Trinomials II
Now let’s try to factor trinomials of the form ax2 + bx + c.
We’ll give two methods. One is short but not reliable.
The second one takes more steps but gives definite answers.
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

17. Factoring Trinomials II
(Reversed FOIL Method)

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

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

22. 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 II
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

23. 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 II
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

24. 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 II
(Reversed FOIL Method)
Let’s see how the above examples are related to factoring.

25. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Factoring Trinomials II
(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,
Factoring Trinomials II
(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,
Factoring Trinomials II
(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. 3(± # ) +1(± #) = 5 where the #’s are 1 and 2.
Since 3(1) +1(2) = 5,
Factoring Trinomials II
(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

29. 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 II
(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

30. 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 II
(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

31. Factoring Trinomials II
Example C. Factor 3x2 – 7x + 2.

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

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

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

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

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

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

44. Example E. Factor 3x2 + 8x + 2.
Factoring Trinomials II

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

46. 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 II

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.
Factoring Trinomials II

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 II

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 II
If both the numbers a and c in ax2 + bx + c have many factors
then there are many possibilities to check.

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.
Factoring Trinomials II
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.

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.
Factoring Trinomials II
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 ± #).

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.
Factoring Trinomials II
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),

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.
Factoring Trinomials II
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

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 II
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.

55. 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 II
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.

56. 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 II
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.

57. 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 II
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).

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

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

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

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 II
It's not necessary to always start with ax2. If c is a prime
number, we start with c.
(± #)(± 3) + (± #)(±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
Factoring Trinomials II
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)

64. 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 II
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.

65. 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 II
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.

66. 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 II
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.

67. 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 II
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.

68. 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 II
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.

69. Example H. Factor 3x2 – 7x – 2 .
Factoring Trinomials II

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

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

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.
Factoring Trinomials II

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 II

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 II
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 II
Example I: Factor 1x2 + 5x – 6 .
There might be multiple matchings for
(#)(± #) + (#)( ± #) = b
make sure you chose the correct one, if any.

76. 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 II
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.

77. 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 II
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.

78. 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 II
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.

79. Factoring Trinomials II
Finally, before starting the reverse-FOIL procedure

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

81. Factoring Trinomials II
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.

82. Factoring Trinomials II
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.

83. Factoring Trinomials II
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

84. Factoring Trinomials II
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

85. Factoring Trinomials II
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)

86. Factoring Trinomials II
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)

87. Factoring Trinomials II
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)

88. 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 II
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