A Clever Method for Factoring Trinomials – Solutions
This simple method of factoring trinomials is what is known in some mathematical circles as “The British
Method of Factoring.” I had been retired from teaching high school and was in my 11th year of conducting
summer mathematics workshops at Bluffton College (now University). One of my participants that year,
Marcia Harper from Cardinal Stritch High School, shared this technique with us in an algebra session, and
all of us were amazed. We had never seen such a neat, straightforward approach to a long, guess-and-check
procedure. Needless to say, I jumped on the technique and have been using it ever since. Thanks, Marcia!!
Procedure:
• Always express the trinomial in the form ax2 + bx + c. So in 2x2 + x – 6, a = 2, b = 1, c = – 6
• Find the absolute value of a times c. |ac| = |2(– 6)| = |–12| = 12
• Find all pairs of factors of |ac|. Look for a Sum or Difference that is b or – b (1 or – 1 in this case).
12
1, 12
2, 6
3, 4
•
•
•
(S) or (D)
13 or 11
8 or 4
7 or 1
NO
NO
YES (4 – 3 = 1)
Rewrite bx as the Sum or Difference. Rewrite 2x2 + x – 6 as 2x2 + 4x – 3x – 6
Factor by grouping. Factor: 2x2 + 4x – 3x – 6 = 2x(x + 2) – 3(x + 2) = (2x – 3)(x + 2)
Multiply the factors to check your work.
Example: Let’s go back to the original problem: 12x2 – 7x – 12.
Find |ac|: |12(–12)| = 144. Find all pairs of factors of 144, looking for a sum or difference of 7 or −7:
144
1, 144
2, 72
3, 48
4, 36
6, 24
8, 18
9, 16
•
•
•
•
(S) or (D)
145 or 143
74 or 70
51 or 45
40 or 32
30 or 18
26 or 10
25 or 7
NO
NO
NO
NO
NO
NO
YES!!!
Rewrite (– 7x) as (9x – 16x) so 12x2 – 7x – 12 = 12x2 + 9x – 16x – 12
Factor by grouping: 12x2 + 9x – 16x – 12 = 3x(4x + 3) – 4(4x + 3) = (3x – 4)(4x + 3)
Multiply the factors to check your work.
Notice that we can also rewrite 12x2 – 7x – 12 as 12x2 – 16x + 9x – 12 = 4x(3x – 4) + 3(3x – 4) =
(4x + 3)(3x – 4) = (3x – 4)(4x + 3), the same answer as above.
Example: Let’s try one more: Factor 6x2 – 7x + 6.
Here |ac| = |6(6)| = 36. Find all pairs of factors of 36 and look for a sum or difference of 7 or − 7.
36
1, 36
2, 18
3, 12
4, 9
6, 6
(S) or (D)
37 or 35
20 or 16
15 or 9
13 or 5
12 or 0
NO
NO
NO
NO
NO
Since neither the sum nor difference of the factors ever equals 7 or – 7, we know that 6x2 – 7x + 6 will
not factor over the set of integers, i.e., it is “prime.”
Practice: Factor completely. (See answers below)
1.
2.
3.
4.
5.
6.
7.
8.
6m2 + 13m + 6
6m2 – 5m – 6
24c2 – 34c – 45
12c2 + 64c + 45
4x2 + 16x – 84
4x2 – 41x – 82
36x2 – 120x + 100
36x2 – 225x – 100
Answers:
1.
6m2 + 13m + 6. Here is a trinomial whose leading coefficient is not 1. Note that |ac| = |6 × 6| = 36.
Find pairs of factors of 36 until their sum or difference is 13:
36
1, 36
2, 18
3, 12
4, 9
Let 13m = 4m + 9m, so 6m2 + 13m + 6 =
=
=
=
(S) or (D)
37 or 35
20 or 16
15 or 9
13 or 5
NO
NO
NO
YES!!
6m2 + 4m + 9m + 6
(6m2 + 4m) + (9m + 6)
2m(3m + 2) + 3(3m + 2)
(2m + 3)(3m + 2)
2. 6m2 – 5m – 6. Note in # 1, we found that the factors 4 and 9 gave us a difference of 5.
This time let the middle term – 5m = 4m – 9m so 6m2 – 5m – 6 =
=
=
=
3.
6m2 + 4m – 9m – 6
(6m2 + 4m) – (9m + 6)
2m(3m + 2) – 3(3m + 2)
(2m – 3)(3m + 2)
24c2 – 34c – 45. Note that |24(-45)| = |–1080| = 1080
1080
1, 1080
2, 540
3, 360
4, 270
5, 216
6, 180
8, 135
9, 120
10, 108
12, 90
15, 72
18, 60
20, 54
(S) or (D)
1081 or 1079
542 or 538
363 or 357
274 or 266
221 or 211
186 or 174
143 or 127
129 or 111
118 or 98
102 or 78
87 or 57
78 or 42
74 or 34
So 24c2 – 34c – 45 = 24c2 + 20c – 54c – 45 = 4c(6c + 5) – 9(6c + 5) = (4c – 9)(6c + 5)
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
FINALLY!!
4.
12c2 + 64c + 45. Note that |12 × 45| = 540. List all pairs of factors of 540 until we find a sum or
difference of 64. In this case we spot right away that 54 and 10 will work.
Thus, 12c2 + 64c + 45 = 12c2 + 54c + 10c + 45 = 6c(2c + 9) + 5(2c + 9) = (6c + 5)(2c + 9)
5.
4x2 + 16x – 84. Note that in this case we can take out a common factor of 4, leaving:
4(x2 + 4x – 21) = 4(x + 7)(x – 3)
6.
4x2 – 41x – 82. Note that |4(−82)| = 328. List all pairs of factors of 328 until we find a sum or
difference of 41.
328
(S) or (D)
1, 328
329 or 327
NO
2, 164
166 or 162
NO
4, 82
86 or 78
NO
8, 41
49 or 33
NO
Since √328 < 19, we have found and tried all possible pairs of factors of 328. Since none of their sums
or differences is equal to 41, we know 4x2 – 41x – 82 is not factorable. It is prime.
Note: We can also use the discriminant (b2 – 4ac) to determine whether a trinomial is factorable over
the set of rational numbers. In this case, b2 – 4ac = (–41)2 – 4(4)(–82) = 2993. Because 2993 is not a
perfect square, 4x2 – 41x – 82 is prime, or not factorable.
7.
36x2 – 120x + 100. Note that there is a common factor of 4:
36x2 – 120x + 100 = 4(9x2 – 30x + 25)
= 4(3x – 5)(3x – 5)
8.
36x2 – 225x – 100. Note that |36(−100)| = 3600.
List all pairs of factors of 3600 until we find a sum or difference of 225.
3600
1, 3600
2, 1800
3, 1200
4, 900
5, 720
6, 600
8, 450
9, 400
10, 360
12, 300
15, 240
Thus we can write:
(S) or (D)
3601 or 3599
1802 or 1798
1203 or 1197
904 or 896
725 or 715
606 or 594
458 or 442
409 or 391
370 or 350
312 or 288
255 or 225
36x2 – 225x – 100 = 36x2 + 15x − 240x – 100
= 3x(12x + 5) − 20(12x + 5)
= (3x − 20)(12x + 5)
NO
NO
NO
NO
NO
NO
NO
NO
NO
NO
YES!!!