SlideShare a Scribd company logo

10 more on factoring trinomials and factoring by formulas

Download as PPTX, PDF
E
elem-alg-sample

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.

Education
1 of 88
Download to read offline
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
Factoring Trinomials II
Factoring Trinomials II Now let’s try to factor trinomials of the form ax2 + bx + c.
Factoring Trinomials II Now let’s try to factor trinomials of the form ax2 + bx + c. We’ll give two methods.
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.
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.
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
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
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
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
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
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
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
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
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
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
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
Factoring Trinomials II (Reversed FOIL Method)
Factoring Trinomials II (Reversed FOIL Method) Let’s see how the above examples are related to factoring.
Example B. Factor 3x2 + 5x + 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 ± #). 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 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. 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. 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, Factoring Trinomials II (Reversed FOIL Method) Let’s see how the above examples are related to factoring.
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
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
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
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
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
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
Factoring Trinomials II Example C. Factor 3x2 – 7x + 2.
Factoring Trinomials II Example C. Factor 3x2 – 7x + 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
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
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
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
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
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
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
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
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
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
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
Example E. Factor 3x2 + 8x + 2. Factoring Trinomials II
Example E. Factor 3x2 + 8x + 2. We start with (3x ± #)(1x ± #). Factoring Trinomials II
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
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
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
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 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.
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 ± #).
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),
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
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.
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.
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.
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).
Factoring Trinomials II It's not necessary to always start with ax2. If c is a prime number, we start with c.
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.
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.
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.
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.
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)
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.
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.
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.
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.
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.
Example H. Factor 3x2 – 7x – 2 . Factoring Trinomials II
Example H. Factor 3x2 – 7x – 2 . We start with (3x ± #)(1x ± #). Factoring Trinomials II
Example H. Factor 3x2 – 7x – 2 . We start with (3x ± #)(1x ± #). We find that: 3(–2) + 1(–1) = –7. Factoring Trinomials II
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
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 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.
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.
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.
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.
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.
Factoring Trinomials II Finally, before starting the reverse-FOIL procedure
Factoring Trinomials II Finally, before starting the reverse-FOIL procedure 1. make sure the terms are arranged in order.
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.
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.
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
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
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)
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)
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)
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

More Related Content

PPTX
5 4factoring trinomial ii
math123a
 
PPTX
5 5factoring trinomial iii
math123a
 
PDF
Exponents and Polynomials
REYBETH RACELIS
 
PDF
rational expressions
TrabahoLang
 
PPTX
2 3linear equations ii
math123a
 
PDF
Chapter0
osamahussien
 
PPTX
4 7polynomial operations-vertical
math123a
 
PDF
Chapter 1 Functions Relations V3
luxvis
 
5 4factoring trinomial ii
math123a
 
5 5factoring trinomial iii
math123a
 
Exponents and Polynomials
REYBETH RACELIS
 
rational expressions
TrabahoLang
 
2 3linear equations ii
math123a
 
Chapter0
osamahussien
 
4 7polynomial operations-vertical
math123a
 
Chapter 1 Functions Relations V3
luxvis
 

What's hot (20)

PDF
123a ppt-all-2
math123a
 
PPT
Functions And Relations
andrewhickson
 
PPTX
Factorising Common Factors
Passy World
 
PPT
Business Math Chapter 3
Nazrin Nazdri
 
PDF
Discrete Mathematics with Applications 4th Edition Susanna Solutions Manual
HayfaVaughan
 
PPT
Business Math Chapter 2
Nazrin Nazdri
 
PDF
Q1 week 1 (common monomial,sum & diffrence of two cubes,difference of tw...
Walden Macabuhay
 
PDF
Factoring Polynomials in Modular Approach
Lorie Jane Letada
 
PPTX
53 multiplication and division of rational expressions
alg1testreview
 
PPTX
5 6 substitution and factoring formulas
math123a
 
PPT
6.2 presentation
Randall Micallef
 
PPTX
Simplifying algebraic expression
julienorman80065
 
PPT
Business Math Chapter 1
Nazrin Nazdri
 
PPTX
Properties of Addition & Multiplication
itutor
 
PPTX
Relations
BRIJESH UPADHYAY
 
PPT
Exponents, Fractions And Factors
nickromero76
 
PPT
9 2power Of Power
taco40
 
PPTX
Unit 7.3
Mark Ryder
 
PPTX
12 rational expressions
elem-alg-sample
 
PPT
Simplifying expressions
Zain Masood
 
123a ppt-all-2
math123a
 
Functions And Relations
andrewhickson
 
Factorising Common Factors
Passy World
 
Business Math Chapter 3
Nazrin Nazdri
 
Discrete Mathematics with Applications 4th Edition Susanna Solutions Manual
HayfaVaughan
 
Business Math Chapter 2
Nazrin Nazdri
 
Q1 week 1 (common monomial,sum & diffrence of two cubes,difference of tw...
Walden Macabuhay
 
Factoring Polynomials in Modular Approach
Lorie Jane Letada
 
53 multiplication and division of rational expressions
alg1testreview
 
5 6 substitution and factoring formulas
math123a
 
6.2 presentation
Randall Micallef
 
Simplifying algebraic expression
julienorman80065
 
Business Math Chapter 1
Nazrin Nazdri
 
Properties of Addition & Multiplication
itutor
 
Relations
BRIJESH UPADHYAY
 
Exponents, Fractions And Factors
nickromero76
 
9 2power Of Power
taco40
 
Unit 7.3
Mark Ryder
 
12 rational expressions
elem-alg-sample
 
Simplifying expressions
Zain Masood
 
Ad

Viewers also liked (10)

PPS
Cute animals
voiculescu gabriel
 
PPS
Friendship Bouquet
Karthik D
 
PPSX
Cute Animals
DINISHA
 
PPS
Beautiful Rose Flowers
Thilini
 
PPT
Factoring Trinomials
guest7370dc
 
PPT
Factoring Trinomials
Don Simmons
 
PPS
Beautiful Thoughts On Friendship
Heather Gill
 
PPT
Factoring ppt 2003 (1)
asheeshkumar09
 
PPT
Factoring ppt
Kapil Chhabra
 
PDF
10 Ice Breaker Games - How to get to know your office
Elodie A.
 
Cute animals
voiculescu gabriel
 
Friendship Bouquet
Karthik D
 
Cute Animals
DINISHA
 
Beautiful Rose Flowers
Thilini
 
Factoring Trinomials
guest7370dc
 
Factoring Trinomials
Don Simmons
 
Beautiful Thoughts On Friendship
Heather Gill
 
Factoring ppt 2003 (1)
asheeshkumar09
 
Factoring ppt
Kapil Chhabra
 
10 Ice Breaker Games - How to get to know your office
Elodie A.
 
Ad

Similar to 10 more on factoring trinomials and factoring by formulas (20)

PDF
Algebra factoring
TrabahoLang
 
PPT
Factoring 15.3 and 15.4 Grouping and Trial and Error
swartzje
 
PPT
P7
salamhello
 
PPTX
Difference of Two Squares
Lorie Jane Letada
 
PPT
P6 factoring
salamhello
 
PPT
P6 factoring
salamhello
 
PDF
3.complex numbers Further Mathematics Zimbabwe Zimsec Cambridge
alproelearning
 
PPTX
1.7 multiplication ii w
Tzenma
 
PPTX
Q1-W1-Factoring Polynomials.pptx
TherezaNoble
 
PPT
Math083 day 1 chapter 6 2013 fall
jbianco9910
 
PDF
Alg complex numbers
TrabahoLang
 
PPT
Chapter p 5
salamhello
 
PPTX
Diapositivas unidad 1
IvnDavidVegaJuris
 
PPTX
factorization for class 8.pptx
SurendraNeupaneItsgo
 
PDF
Quadratic Equations
Wenslette Rosique
 
PPT
Factoring polynomials
Mark Ryder
 
PPT
Swartz Factoring
swartzje
 
PDF
Maths paper class 12 maths paper class 12
nsridevi1405
 
PPTX
Paso 2 profundizar y contextualizar el conocimiento de la unidad 1 ff
ALGEBRAGEOMETRIA
 
PPT
Perfect Squares
guest02fe304
 
Algebra factoring
TrabahoLang
 
Factoring 15.3 and 15.4 Grouping and Trial and Error
swartzje
 
P7
salamhello
 
Difference of Two Squares
Lorie Jane Letada
 
P6 factoring
salamhello
 
P6 factoring
salamhello
 
3.complex numbers Further Mathematics Zimbabwe Zimsec Cambridge
alproelearning
 
1.7 multiplication ii w
Tzenma
 
Q1-W1-Factoring Polynomials.pptx
TherezaNoble
 
Math083 day 1 chapter 6 2013 fall
jbianco9910
 
Alg complex numbers
TrabahoLang
 
Chapter p 5
salamhello
 
Diapositivas unidad 1
IvnDavidVegaJuris
 
factorization for class 8.pptx
SurendraNeupaneItsgo
 
Quadratic Equations
Wenslette Rosique
 
Factoring polynomials
Mark Ryder
 
Swartz Factoring
swartzje
 
Maths paper class 12 maths paper class 12
nsridevi1405
 
Paso 2 profundizar y contextualizar el conocimiento de la unidad 1 ff
ALGEBRAGEOMETRIA
 
Perfect Squares
guest02fe304
 

More from elem-alg-sample (20)

PPTX
6 equations and applications of lines
elem-alg-sample
 
PPTX
5 slopes of lines
elem-alg-sample
 
PPTX
4 linear equations and graphs of lines
elem-alg-sample
 
PPTX
3 rectangular coordinate system
elem-alg-sample
 
PPTX
2 the real line, inequalities and comparative phrases
elem-alg-sample
 
PPTX
1 basic geometry and formulas
elem-alg-sample
 
PPTX
18 variations
elem-alg-sample
 
PPTX
17 applications of proportions and the rational equations
elem-alg-sample
 
PPTX
16 the multiplier method for simplifying complex fractions
elem-alg-sample
 
PPTX
15 proportions and the multiplier method for solving rational equations
elem-alg-sample
 
PPTX
14 the lcm and the multiplier method for addition and subtraction of rational...
elem-alg-sample
 
PPTX
13 multiplication and division of rational expressions
elem-alg-sample
 
PPTX
11 applications of factoring
elem-alg-sample
 
PPTX
9 factoring trinomials
elem-alg-sample
 
PPTX
8 factoring out gcf
elem-alg-sample
 
PPTX
7 special binomial operations and formulas
elem-alg-sample
 
PPTX
6 polynomial expressions and operations
elem-alg-sample
 
PPTX
5 exponents and scientific notation
elem-alg-sample
 
PPTX
4 literal equations
elem-alg-sample
 
PPTX
3 linear equations
elem-alg-sample
 
6 equations and applications of lines
elem-alg-sample
 
5 slopes of lines
elem-alg-sample
 
4 linear equations and graphs of lines
elem-alg-sample
 
3 rectangular coordinate system
elem-alg-sample
 
2 the real line, inequalities and comparative phrases
elem-alg-sample
 
1 basic geometry and formulas
elem-alg-sample
 
18 variations
elem-alg-sample
 
17 applications of proportions and the rational equations
elem-alg-sample
 
16 the multiplier method for simplifying complex fractions
elem-alg-sample
 
15 proportions and the multiplier method for solving rational equations
elem-alg-sample
 
14 the lcm and the multiplier method for addition and subtraction of rational...
elem-alg-sample
 
13 multiplication and division of rational expressions
elem-alg-sample
 
11 applications of factoring
elem-alg-sample
 
9 factoring trinomials
elem-alg-sample
 
8 factoring out gcf
elem-alg-sample
 
7 special binomial operations and formulas
elem-alg-sample
 
6 polynomial expressions and operations
elem-alg-sample
 
5 exponents and scientific notation
elem-alg-sample
 
4 literal equations
elem-alg-sample
 
3 linear equations
elem-alg-sample
 

Recently uploaded (20)

PDF
Anesthesia in Laparoscopic Surgery in India
mohitsuren827
 
PDF
Computing-Curriculum for Schools in Ghana
abigailanane203
 
PPTX
IMMUNITY IMMUNITY refers to protection against infection, and the immune syst...
shilpa939953
 
PPTX
school management -TNTEU- B.Ed., Semester II Unit 1.pptx
NihumathunnisaH
 
PPTX
Cell Structure & Organelles in detailed.
DHANASHREESIVAKUMAR
 
PPTX
Pharma ospi slides which help in ospi learning
rameetkumar304
 
PPTX
1st Inaugural Professorial Lecture held on 19th February 2020 (Governance and...
kawisojames10
 
PDF
3rd Neelam Sanjeevareddy Memorial Lecture.pdf
Academy of Grassroots Studies and Research of India (AGRASRI)
 
PDF
Insiders guide to clinical Medicine.pdf
AbhishekPatil315128
 
PPTX
Renaissance Architecture: A Journey from Faith to Humanism
DrMonicaSharma
 
PDF
Supply Chain Operations Speaking Notes -ICLT Program
labyankof
 
PDF
FourierSeries-QuestionsWithAnswers(Part-A).pdf
Raji Murugan
 
PDF
Abdominal Access Techniques with Prof. Dr. R K Mishra
mohitsuren827
 
PDF
Black Hat USA 2025 - Micro ICS Summit - ICS/OT Threat Landscape
Chris Sistrunk
 
PPTX
GDM (1) (1).pptx small presentation for students
shrawanbpkihs
 
PPTX
Pharmacology of Heart Failure /Pharmacotherapy of CHF
Rajshri Ghogare
 
PDF
Chapter 2 Heredity, Prenatal Development, and Birth.pdf
MarjorieLopezTiu
 
PPTX
Microbial diseases, their pathogenesis and prophylaxis
DevlinaSengupta
 
PDF
O5-L3 Freight Transport Ops (International) V1.pdf
labyankof
 
PDF
STATICS OF THE RIGID BODIES Hibbelers.pdf
pascualasturiaskaye4
 
Anesthesia in Laparoscopic Surgery in India
mohitsuren827
 
Computing-Curriculum for Schools in Ghana
abigailanane203
 
IMMUNITY IMMUNITY refers to protection against infection, and the immune syst...
shilpa939953
 
school management -TNTEU- B.Ed., Semester II Unit 1.pptx
NihumathunnisaH
 
Cell Structure & Organelles in detailed.
DHANASHREESIVAKUMAR
 
Pharma ospi slides which help in ospi learning
rameetkumar304
 
1st Inaugural Professorial Lecture held on 19th February 2020 (Governance and...
kawisojames10
 
3rd Neelam Sanjeevareddy Memorial Lecture.pdf
Academy of Grassroots Studies and Research of India (AGRASRI)
 
Insiders guide to clinical Medicine.pdf
AbhishekPatil315128
 
Renaissance Architecture: A Journey from Faith to Humanism
DrMonicaSharma
 
Supply Chain Operations Speaking Notes -ICLT Program
labyankof
 
FourierSeries-QuestionsWithAnswers(Part-A).pdf
Raji Murugan
 
Abdominal Access Techniques with Prof. Dr. R K Mishra
mohitsuren827
 
Black Hat USA 2025 - Micro ICS Summit - ICS/OT Threat Landscape
Chris Sistrunk
 
GDM (1) (1).pptx small presentation for students
shrawanbpkihs
 
Pharmacology of Heart Failure /Pharmacotherapy of CHF
Rajshri Ghogare
 
Chapter 2 Heredity, Prenatal Development, and Birth.pdf
MarjorieLopezTiu
 
Microbial diseases, their pathogenesis and prophylaxis
DevlinaSengupta
 
O5-L3 Freight Transport Ops (International) V1.pdf
labyankof
 
STATICS OF THE RIGID BODIES Hibbelers.pdf
pascualasturiaskaye4
 

10 more on factoring trinomials and factoring by formulas

  • 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
  • 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