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
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
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
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).
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.
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.
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)