Free Printable Factoring Trinomials Worksheets - Free Printable

You’ve uploaded a worksheet titled “Factoring Trinomials (a > 1)” from Kuta Software - Infinite Algebra 1. The goal is to factor each trinomial completely using the “split the middle term” method (also called factoring by grouping), which is standard for trinomials of the form `ax² + bx + c` where `a > 1`.

Let’s go through each problem, verify the handwritten answers, and explain the correct factoring process step-by-step.

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✔ Problem 1: `3p² - 2p - 5`

Handwritten Answer: `(3p - 5)(p + 1)` — ✔️ Correct

Explanation:
- Multiply `a * c = 3 * (-5) = -15`
- Find two numbers that multiply to `-15` and add to `-2` → 3 and -5
- Split middle term:
`3p² + 3p - 5p - 5`
- Group:
`(3p² + 3p) + (-5p - 5)`
- Factor each group:
`3p(p + 1) - 5(p + 1)`
- Factor out common binomial:
`(3p - 5)(p + 1)`

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✔ Problem 2: `2n² + 3n - 9`

Handwritten Answer: `(2n - 3)(n + 3)` — ✔️ Correct

Explanation:
- `a * c = 2 * (-9) = -18`
- Two numbers that multiply to `-18`, add to `+3` → 6 and -3
- Split: `2n² + 6n - 3n - 9`
- Group: `(2n² + 6n) + (-3n - 9)`
- Factor: `2n(n + 3) - 3(n + 3)`
- Final: `(2n - 3)(n + 3)`

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✔ Problem 3: `3n² - 8n + 4`

Handwritten Answer: `(3n - 2)(n - 2)` — ✔️ Correct

Explanation:
- `a * c = 3 * 4 = 12`
- Two numbers that multiply to `12`, add to `-8` → -6 and -2
- Split: `3n² - 6n - 2n + 4`
- Group: `(3n² - 6n) + (-2n + 4)`
- Factor: `3n(n - 2) - 2(n - 2)`
- Final: `(3n - 2)(n - 2)`

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✔ Problem 4: `5n² + 19n + 12`

Handwritten Answer: `(5n + 4)(n + 3)` — ✔️ Correct

Explanation:
- `a * c = 5 * 12 = 60`
- Two numbers that multiply to `60`, add to `19` → 15 and 4
- Split: `5n² + 15n + 4n + 12`
- Group: `(5n² + 15n) + (4n + 12)`
- Factor: `5n(n + 3) + 4(n + 3)`
- Final: `(5n + 4)(n + 3)`

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✔ Problem 5: `2v² + 11v + 5`

Handwritten Answer: `(2v + 1)(v + 5)` — ✔️ Correct

Explanation:
- `a * c = 2 * 5 = 10`
- Two numbers that multiply to `10`, add to `11` → 10 and 1
- Split: `2v² + 10v + v + 5`
- Group: `(2v² + 10v) + (v + 5)`
- Factor: `2v(v + 5) + 1(v + 5)`
- Final: `(2v + 1)(v + 5)`

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✔ Problem 6: `2n² + 5n + 2`

Handwritten Answer: `(2n + 1)(n + 2)` — ✔️ Correct

Explanation:
- `a * c = 2 * 2 = 4`
- Two numbers that multiply to `4`, add to `5` → 4 and 1
- Split: `2n² + 4n + n + 2`
- Group: `(2n² + 4n) + (n + 2)`
- Factor: `2n(n + 2) + 1(n + 2)`
- Final: `(2n + 1)(n + 2)`

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✔ Problem 7: `7a² + 53a + 28`

Handwritten Answer: `(7a + 4)(a + 7)` — ✔️ Correct

Explanation:
- `a * c = 7 * 28 = 196`
- Two numbers that multiply to `196`, add to `53` → 49 and 4
- Split: `7a² + 49a + 4a + 28`
- Group: `(7a² + 49a) + (4a + 28)`
- Factor: `7a(a + 7) + 4(a + 7)`
- Final: `(7a + 4)(a + 7)`

*(Note: The student wrote “(7a+4)(a+1)” in one line but corrected it to “(7a+4)(a+7)” — likely a typo. Final answer is correct.)*

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✔ Problem 8: `9k² + 66k + 21`

Handwritten Answer: `3(3k + 1)(k + 7)` — ✔️ Correct

Explanation:
- First, factor out GCF: All terms divisible by 3 →
`3(3k² + 22k + 7)`
- Now factor `3k² + 22k + 7`
- `a * c = 3 * 7 = 21`
- Two numbers that multiply to `21`, add to `22` → 21 and 1
- Split: `3k² + 21k + k + 7`
- Group: `(3k² + 21k) + (k + 7)`
- Factor: `3k(k + 7) + 1(k + 7)`
- Final: `(3k + 1)(k + 7)`
- Don’t forget the GCF:
`3(3k + 1)(k + 7)`

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## 🎯 Summary

✔ All 8 problems are correctly factored.

The student used the AC method (splitting the middle term) effectively for each problem, then grouped and factored out the common binomial. In Problem 8, they also correctly identified and factored out the GCF first, which is essential for "factoring completely."

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## 💡 Tips for Success:

- Always check if there’s a GCF before splitting the middle term.
- Double-check your signs — especially when dealing with negative constants.
- After factoring, you can multiply back to verify your answer matches the original trinomial.

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✔ Final Grade: A+ — Excellent work! All solutions are accurate and well-explained in the steps shown.