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To determine if Lucas or Erick correctly grouped and factored the polynomial [tex]\(12x^3 - 6x^2 + 8x - 4\)[/tex], let's carefully analyze their steps and validate their factorization.
### Lucas’s Grouping and Factoring
Lucas grouped the polynomial as:
[tex]\[ (12x^3 + 8x) + (-6x^2 - 4) \][/tex]
Step-by-Step Analysis:
1. Group 1: [tex]\(12x^3 + 8x\)[/tex]
Factor out the common term [tex]\(4x\)[/tex]:
[tex]\[ 12x^3 + 8x = 4x(3x^2 + 2) \][/tex]
2. Group 2: [tex]\(-6x^2 - 4\)[/tex]
Factor out the common term [tex]\(-2\)[/tex]:
[tex]\[ -6x^2 - 4 = -2(3x^2 + 2) \][/tex]
After factoring, we combine the factored groups:
[tex]\[ 4x(3x^2 + 2) - 2(3x^2 + 2) \][/tex]
Since [tex]\((3x^2 + 2)\)[/tex] is a common factor, we can further factorize:
[tex]\[ = (3x^2 + 2)(4x - 2) \][/tex]
### Erick’s Grouping and Factoring
Erick grouped the polynomial as:
[tex]\[ (12x^3 - 6x^2) + (8x - 4) \][/tex]
Step-by-Step Analysis:
1. Group 1: [tex]\(12x^3 - 6x^2\)[/tex]
Factor out the common term [tex]\(6x^2\)[/tex]:
[tex]\[ 12x^3 - 6x^2 = 6x^2(2x - 1) \][/tex]
2. Group 2: [tex]\(8x - 4\)[/tex]
Factor out the common term [tex]\(4\)[/tex]:
[tex]\[ 8x - 4 = 4(2x - 1) \][/tex]
After factoring, we combine the factored groups:
[tex]\[ 6x^2(2x - 1) + 4(2x - 1) \][/tex]
Since [tex]\((2x - 1)\)[/tex] is a common factor, we can further factorize:
[tex]\[ = (2x - 1)(6x^2 + 4) \][/tex]
### Conclusion
Both Lucas and Erick correctly grouped the terms for factoring and proceeded with correct factorization steps. However, the method of grouping affects the intermediate factorizations, yet both should arrive at an equivalent polynomial.
To sum up:
- Lucas’s groupings and factorings:
[tex]\[ (12x^3 + 8x) + (-6x^2 - 4) = 4x(3x^2 + 2) - 2(3x^2 + 2) = (3x^2 + 2)(4x - 2) \][/tex]
- Erick’s groupings and factorings:
[tex]\[ (12x^3 - 6x^2) + (8x - 4) = 6x^2(2x - 1) + 4(2x - 1) = (2x - 1)(6x^2 + 4) \][/tex]
So, both Lucas and Erick have used correct grouping to factor the polynomial [tex]\(12x^3 - 6x^2 + 8x - 4\)[/tex]. Both approaches are valid, and the factorizations are both correct and equivalent in their results.
### Lucas’s Grouping and Factoring
Lucas grouped the polynomial as:
[tex]\[ (12x^3 + 8x) + (-6x^2 - 4) \][/tex]
Step-by-Step Analysis:
1. Group 1: [tex]\(12x^3 + 8x\)[/tex]
Factor out the common term [tex]\(4x\)[/tex]:
[tex]\[ 12x^3 + 8x = 4x(3x^2 + 2) \][/tex]
2. Group 2: [tex]\(-6x^2 - 4\)[/tex]
Factor out the common term [tex]\(-2\)[/tex]:
[tex]\[ -6x^2 - 4 = -2(3x^2 + 2) \][/tex]
After factoring, we combine the factored groups:
[tex]\[ 4x(3x^2 + 2) - 2(3x^2 + 2) \][/tex]
Since [tex]\((3x^2 + 2)\)[/tex] is a common factor, we can further factorize:
[tex]\[ = (3x^2 + 2)(4x - 2) \][/tex]
### Erick’s Grouping and Factoring
Erick grouped the polynomial as:
[tex]\[ (12x^3 - 6x^2) + (8x - 4) \][/tex]
Step-by-Step Analysis:
1. Group 1: [tex]\(12x^3 - 6x^2\)[/tex]
Factor out the common term [tex]\(6x^2\)[/tex]:
[tex]\[ 12x^3 - 6x^2 = 6x^2(2x - 1) \][/tex]
2. Group 2: [tex]\(8x - 4\)[/tex]
Factor out the common term [tex]\(4\)[/tex]:
[tex]\[ 8x - 4 = 4(2x - 1) \][/tex]
After factoring, we combine the factored groups:
[tex]\[ 6x^2(2x - 1) + 4(2x - 1) \][/tex]
Since [tex]\((2x - 1)\)[/tex] is a common factor, we can further factorize:
[tex]\[ = (2x - 1)(6x^2 + 4) \][/tex]
### Conclusion
Both Lucas and Erick correctly grouped the terms for factoring and proceeded with correct factorization steps. However, the method of grouping affects the intermediate factorizations, yet both should arrive at an equivalent polynomial.
To sum up:
- Lucas’s groupings and factorings:
[tex]\[ (12x^3 + 8x) + (-6x^2 - 4) = 4x(3x^2 + 2) - 2(3x^2 + 2) = (3x^2 + 2)(4x - 2) \][/tex]
- Erick’s groupings and factorings:
[tex]\[ (12x^3 - 6x^2) + (8x - 4) = 6x^2(2x - 1) + 4(2x - 1) = (2x - 1)(6x^2 + 4) \][/tex]
So, both Lucas and Erick have used correct grouping to factor the polynomial [tex]\(12x^3 - 6x^2 + 8x - 4\)[/tex]. Both approaches are valid, and the factorizations are both correct and equivalent in their results.
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