Algebra 1 Practice - Factor by Grouping (Example 1)

In **Algebra 1**, factoring by **grouping** is used when there are four terms in a polynomial. The goal is to group terms and factor each group, then factor the entire expression.

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### **Steps to Factor by Grouping**

1. **Group the Terms:**
- Divide the polynomial into two groups.

2. **Factor Each Group:**
- Factor out the GCF from each group.

3. **Find a Common Binomial Factor:**
- If the groups share a common binomial factor, factor it out.

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### **Examples**

1. **Factor \( x^3 + 3x^2 + 2x + 6 \):**
- Group terms: \( (x^3 + 3x^2) + (2x + 6) \).
- Factor each group: \( x^2(x + 3) + 2(x + 3) \).
- Factor the binomial: \( (x^2 + 2)(x + 3) \).
- **Answer**: \( (x^2 + 2)(x + 3) \).

2. **Factor \( 6x^3 - 9x^2 + 4x - 6 \):**
- Group terms: \( (6x^3 - 9x^2) + (4x - 6) \).
- Factor each group: \( 3x^2(2x - 3) + 2(2x - 3) \).
- Factor the binomial: \( (3x^2 + 2)(2x - 3) \).
- **Answer**: \( (3x^2 + 2)(2x - 3) \).

3. **Factor \( x^3 - x^2 + 5x - 5 \):**
- Group terms: \( (x^3 - x^2) + (5x - 5) \).
- Factor each group: \( x^2(x - 1) + 5(x - 1) \).
- Factor the binomial: \( (x^2 + 5)(x - 1) \).
- **Answer**: \( (x^2 + 5)(x - 1) \).

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### **Practice Problems**

1. Factor \( x^3 + 4x^2 + 2x + 8 \).
2. Factor \( 2x^3 - 6x^2 + x - 3 \).
3. Factor \( 3x^3 - 9x^2 + 4x - 12 \).

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Nick Perich
Norristown Area High School
Norristown Area School District
Norristown, Pa

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