Handwritten solutions to algebra factoring by grouping problems on a worksheet.

You've uploaded a worksheet titled “Factoring By Grouping” from *Infinite Algebra 2*. The task is to factor each polynomial completely using the method of grouping.

Let’s go through each problem step by step, verify the given answers, and explain the reasoning clearly.

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## 🔢 1) \( 12a^3 - 9a^2 + 4a - 3 \)

✔ Given Answer: \( (4a - 3)(3a^2 + 1) \)

✔ Step-by-step:

Group terms:
\[
(12a^3 - 9a^2) + (4a - 3)
\]

Factor out GCF from each group:
- From first group: \( 3a^2(4a - 3) \)
- From second group: \( 1(4a - 3) \)

Now you have:
\[
3a^2(4a - 3) + 1(4a - 3)
\]

Factor out the common binomial \( (4a - 3) \):
\[
(4a - 3)(3a^2 + 1)
\]

✔️ Correct!

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## 🔢 2) \( 2p^3 + 5p^2 + 6p + 15 \)

✔ Given Answer: \( (2p + 5)(p^2 + 3) \)

✔ Step-by-step:

Group terms:
\[
(2p^3 + 5p^2) + (6p + 15)
\]

Factor out GCF from each group:
- First group: \( p^2(2p + 5) \)
- Second group: \( 3(2p + 5) \)

Now you have:
\[
p^2(2p + 5) + 3(2p + 5)
\]

Factor out common binomial \( (2p + 5) \):
\[
(2p + 5)(p^2 + 3)
\]

✔️ Correct!

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## 🔢 3) \( 3n^3 - 4n^2 + 9n - 12 \)

✔ Given Answer: \( (n^2 + 3)(3n - 4) \)

✔ Step-by-step:

Group terms:
\[
(3n^3 - 4n^2) + (9n - 12)
\]

Factor out GCF from each group:
- First group: \( n^2(3n - 4) \)
- Second group: \( 3(3n - 4) \)

Now you have:
\[
n^2(3n - 4) + 3(3n - 4)
\]

Factor out common binomial \( (3n - 4) \):
\[
(3n - 4)(n^2 + 3)
\]

Note: Order doesn’t matter — \( (n^2 + 3)(3n - 4) \) is the same.

✔️ Correct!

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## 🔢 4) \( 12n^3 + 4n^2 + 3n + 1 \)

✔ Given Answer: \( (4n^2 + 1)(3n + 1) \)

✔ Step-by-step:

Group terms:
\[
(12n^3 + 4n^2) + (3n + 1)
\]

Factor out GCF from each group:
- First group: \( 4n^2(3n + 1) \)
- Second group: \( 1(3n + 1) \)

Now you have:
\[
4n^2(3n + 1) + 1(3n + 1)
\]

Factor out common binomial \( (3n + 1) \):
\[
(3n + 1)(4n^2 + 1)
\]

Same as given answer — just reordered.

✔️ Correct!

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## 🔢 5) \( m^3 - m^2 + 2m - 2 \)

✔ Given Answer: \( (m^2 + 2)(m - 1) \)

✔ Step-by-step:

Group terms:
\[
(m^3 - m^2) + (2m - 2)
\]

Factor out GCF from each group:
- First group: \( m^2(m - 1) \)
- Second group: \( 2(m - 1) \)

Now you have:
\[
m^2(m - 1) + 2(m - 1)
\]

Factor out common binomial \( (m - 1) \):
\[
(m - 1)(m^2 + 2)
\]

Same as given answer — reordered.

✔️ Correct!

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## 🔢 6) \( 5n^3 - 10n^2 + 3n - 6 \)

✔ Given Answer: \( (5n^2 + 3)(n - 2) \)

✔ Step-by-step:

Group terms:
\[
(5n^3 - 10n^2) + (3n - 6)
\]

Factor out GCF from each group:
- First group: \( 5n^2(n - 2) \)
- Second group: \( 3(n - 2) \)

Now you have:
\[
5n^2(n - 2) + 3(n - 2)
\]

Factor out common binomial \( (n - 2) \):
\[
(n - 2)(5n^2 + 3)
\]

Same as given answer — reordered.

✔️ Correct!

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## 🧠 Summary: What is “Factoring by Grouping”?

This method works for polynomials with 4 terms. Steps:

1. Group into two pairs.
2. Factor out the GCF from each pair.
3. If the resulting binomials are the same, factor that binomial out.
4. You now have a product of two factors.

It’s essentially reversing the distributive property:
\( A(C) + B(C) = (A + B)(C) \)

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## ✔ Final Verdict:

All six problems are correctly factored!

The student has demonstrated a solid understanding of factoring by grouping. Each solution follows the correct procedure, and the final factored forms are accurate.

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If you’d like me to explain any specific step in more detail or show how to check your answers (by expanding), feel free to ask!