Question
Factoring By Grouping Date \( \qquad \) Period Factor each completely. 1) \( 12 a^{3}-9 a^{2}+4 a-3 \) 2) \( 2 p^{3}+5 p^{2}+6 p+15 \) 3) \( 3 n^{3}-4 n^{2}+9 n-12 \) 4) \( 12 n^{3}+4 n^{2}+3 n+1 \) 5) \( m^{3}-m^{2}+2 m-2 \) 6) \( 5 n^{3}-10 n^{2}+3 n-6 \) 7) \( 35 x y-5 x-56 y+8 \) 8) \( 224 a z+56 a c-84 y z-21 y c \)
Ask by Edwards Pearson. in South Africa
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
1. \( (3a^{2} + 1)(4a - 3) \)
2. \( (p^{2} + 3)(2p + 5) \)
3. \( (n^{2} + 3)(3n - 4) \)
4. \( (4n^{2} + 1)(3n + 1) \)
5. \( (m^{2} + 2)(m - 1) \)
6. \( (n - 2)(5n^{2} + 3) \)
7. \( (5x - 8)(7y - 1) \)
8. \( 7(8a - 3y)(4z + c) \)
Solution
Sure, let's factor each polynomial by grouping step by step.
---
### **1) \( 12a^{3} - 9a^{2} + 4a - 3 \)**
**Step-by-Step:**
1. **Group terms:**
\((12a^3 - 9a^2) + (4a - 3)\)
2. **Factor out the greatest common factor (GCF) from each group:**
\(3a^2(4a - 3) + 1(4a - 3)\)
3. **Factor out the common binomial:**
\((3a^2 + 1)(4a - 3)\)
**Factored Form:**
\[
\boxed{(3a^{2} + 1)(4a - 3)}
\]
---
### **2) \( 2p^{3} + 5p^{2} + 6p + 15 \)**
**Step-by-Step:**
1. **Group terms:**
\((2p^3 + 5p^2) + (6p + 15)\)
2. **Factor out the GCF from each group:**
\(p^2(2p + 5) + 3(2p + 5)\)
3. **Factor out the common binomial:**
\((p^2 + 3)(2p + 5)\)
**Factored Form:**
\[
\boxed{(p^{2} + 3)(2p + 5)}
\]
---
### **3) \( 3n^{3} - 4n^{2} + 9n - 12 \)**
**Step-by-Step:**
1. **Group terms:**
\((3n^3 - 4n^2) + (9n - 12)\)
2. **Factor out the GCF from each group:**
\(n^2(3n - 4) + 3(3n - 4)\)
3. **Factor out the common binomial:**
\((n^2 + 3)(3n - 4)\)
**Factored Form:**
\[
\boxed{(n^{2} + 3)(3n - 4)}
\]
---
### **4) \( 12n^{3} + 4n^{2} + 3n + 1 \)**
**Step-by-Step:**
1. **Group terms:**
\((12n^3 + 4n^2) + (3n + 1)\)
2. **Factor out the GCF from each group:**
\(4n^2(3n + 1) + 1(3n + 1)\)
3. **Factor out the common binomial:**
\((4n^2 + 1)(3n + 1)\)
**Factored Form:**
\[
\boxed{(4n^{2} + 1)(3n + 1)}
\]
---
### **5) \( m^{3} - m^{2} + 2m - 2 \)**
**Step-by-Step:**
1. **Group terms:**
\((m^3 - m^2) + (2m - 2)\)
2. **Factor out the GCF from each group:**
\(m^2(m - 1) + 2(m - 1)\)
3. **Factor out the common binomial:**
\((m^2 + 2)(m - 1)\)
**Factored Form:**
\[
\boxed{(m^{2} + 2)(m - 1)}
\]
---
### **6) \( 5n^{3} - 10n^{2} + 3n - 6 \)**
**Step-by-Step:**
1. **Group terms:**
\((5n^3 - 10n^2) + (3n - 6)\)
2. **Factor out the GCF from each group:**
\(5n^2(n - 2) + 3(n - 2)\)
3. **Factor out the common binomial:**
\((5n^2 + 3)(n - 2)\)
**Factored Form:**
\[
\boxed{(n - 2)(5n^{2} + 3)}
\]
---
### **7) \( 35xy - 5x - 56yz + 8 \)**
**Step-by-Step:**
1. **Group terms:**
\((35xy - 5x) + (-56yz + 8)\)
2. **Factor out the GCF from each group:**
\(5x(7y - 1) - 8(7y - 1)\)
3. **Factor out the common binomial:**
\((5x - 8)(7y - 1)\)
**Factored Form:**
\[
\boxed{(5x - 8)(7y - 1)}
\]
---
### **8) \( 224az + 56ac - 84yz - 21yc \)**
**Step-by-Step:**
1. **Group terms:**
\((224az + 56ac) + (-84yz - 21yc)\)
2. **Factor out the GCF from each group:**
\(56a(4z + c) - 21y(4z + c)\)
3. **Factor out the common binomial:**
\((56a - 21y)(4z + c)\)
4. **Factor out the GCF from the first binomial:**
\(7(8a - 3y)(4z + c)\)
**Factored Form:**
\[
\boxed{7(8a - 3y)(4z + c)}
\]
---
Each polynomial has been successfully factored by grouping, resulting in the expressions above.
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Bonus Knowledge
Factoring by grouping is a method that cleverly chunks polynomials into smaller parts to identify common factors. It’s like solving a puzzle! For example, if you take the polynomial \( 12a^3 - 9a^2 + 4a - 3 \), you can rearrange and group the terms: \( (12a^3 - 9a^2) + (4a - 3) \). This reveals common factors in each group, leading to a simplified expression! Have you ever tried factoring by grouping in real-world scenarios? It’s not just for math problems in textbooks! For example, if you’re splitting costs among friends for a group outing or organizing your expenses, factoring can help in identifying common expenses (or savings!) across different categories. This approach can make budgeting much simpler and more efficient!
