5.3 Factoring Trinomials (a ≠ 1)
Learning Objectives
By the end of this section, you should be able to:
- Factor trinomials of the form ax² + bx + c where a ≠ 1.
- Use the ac-method (grouping) to factor trinomials.
- Use the trial-and-error (FOIL) method as an alternative strategy.
- Factor out a GCF before factoring when needed.
- Identify prime trinomials.
- Verify factorizations by expanding.
Recognizing the Challenge
When the leading coefficient a ≠ 1, such as in 6x² + 7x + 2, you must use a different strategy.
Instead of multiplying to c, find two numbers that:
- Multiply to a × c
- Add to b
Then rewrite the middle term and factor by grouping.
Factoring by Grouping (ac-Method)
Steps
- Multiply a × c.
- Find two integers that multiply to a × c and add to b.
- Rewrite the middle term using those integers.
- Factor by grouping.
- If no such integers exist, the trinomial is prime.
Example 5.3.1
Factor: 6x² + 7x + 2
Solution
a × c = 12
3 + 4 = 7
6x² + 3x + 4x + 2
(6x² + 3x) + (4x + 2)
3x(2x + 1) + 2(2x + 1)
(3x + 2)(2x + 1)
Example 5.3.2
Factor: 2x² – 7x + 3
Solution
a × c = 6
–6 + –1 = –7
2x² – 6x – x + 3
(2x² – 6x) – (x – 3)
2x(x – 3) – 1(x – 3)
(2x – 1)(x – 3)
Example 5.3.3
Factor: 3x² + 5x – 2
Solution
a × c = –6
6 + (–1) = 5
3x² + 6x – x – 2
(3x² + 6x) – (x + 2)
3x(x + 2) – 1(x + 2)
(3x – 1)(x + 2)
Example 5.3.4
Factor: 3x² + 6x + 4
Solution
a × c = 12
No factor pair of 12 adds to 6
Prime Polynomial
Alternative Method: Trial and Error (FOIL)
Example 5.3.5
Factor: 3x² + 5x – 2
Solution
Try (3x – 1)(x + 2):
3x² + 6x – x – 2 = 3x² + 5x – 2
(3x – 1)(x + 2)
Example 5.3.6
Factor: 4x² – 13x – 12
Solution
ac-method: a × c = –48 → –16 and 3
4x² – 16x + 3x – 12
(4x² – 16x) + (3x – 12)
4x(x – 4) + 3(x – 4)
(4x + 3)(x – 4)
Trial and error confirms the same result.
Try It
When a GCF Appears First
Example 5.3.7
Factor: 9x² + 21x + 10
Solution
a × c = 90 → 15 + 6 = 21
9x² + 15x + 6x + 10
(9x² + 15x) + (6x + 10)
3x(3x + 5) + 2(3x + 5)
(3x + 2)(3x + 5)
Example 5.3.8
Factor: 2x³ + 8x² + 6x
Solution
GCF = 2x
2x(x² + 4x + 3)
2x(x + 1)(x + 3)
Try It
Multivariable Trinomials (a ≠ 1)
Example 5.3.9
Factor: 2a² + 7ab + 3b²
Solution
a × c = 6b² → 6 + 1 = 7
2a² + 6ab + ab + 3b²
(2a² + 6ab) + (ab + 3b²)
(2a + b)(a + 3b)
Example 5.3.10
Factor: 3a² – 5ab – 2b²
Solution
a × c = –6b² → –6 + 1 = –5
3a² – 6ab + ab – 2b²
(3a² – 6ab) + (ab – 2b²)
(3a + b)(a – 2b)
Example 5.3.11
Factor: 4a² + 2ab + 3b²
Solution
a × c = 12b²
No factor pair adds to 2b
This is a Prime Polynomial
Try It
Common Mistakes
🚩 Forgetting to multiply a × c.
🚩 Skipping the GCF step.
🚩 Incorrect signs during grouping.
🚩 Declaring a trinomial prime too quickly.
Key Takeaways
- Multiply a × c, then find integers that multiply to a × c and add to b.
- Split the middle term and factor by grouping.
- Always factor out the GCF first.
- Trial-and-error works for small coefficients.
- Verify by expanding.
