Certainly! Let's analyze the factoring approaches of both Lucas and Erick step by step.
### Lucas' Method:
Lucas groups the polynomial as:
[tex]\[
(12x^3 + 8x) + (-6x^2 - 4)
\][/tex]
1. First Group: [tex]\( 12x^3 + 8x \)[/tex]
- Identify common factors:
[tex]\[
12x^3 + 8x = 4x(3x^2 + 2)
\][/tex]
2. Second Group: [tex]\( -6x^2 - 4 \)[/tex]
- Identify common factors:
[tex]\[
-6x^2 - 4 = -2(3x^2 + 2)
\][/tex]
3. Combine Factored Terms:
- Combine the factored forms within each group:
[tex]\[
4x(3x^2 + 2) + (-2)(3x^2 + 2)
\][/tex]
- Factor out the common binomial term [tex]\( (3x^2 + 2) \)[/tex]:
[tex]\[
(3x^2 + 2)(4x - 2)
\][/tex]
### Erick's Method:
Erick groups the polynomial as:
[tex]\[
(12x^3 - 6x^2) + (8x - 4)
\][/tex]
1. First Group: [tex]\( 12x^3 - 6x^2 \)[/tex]
- Identify common factors:
[tex]\[
12x^3 - 6x^2 = 6x^2(2x - 1)
\][/tex]
2. Second Group: [tex]\( 8x - 4 \)[/tex]
- Identify common factors:
[tex]\[
8x - 4 = 4(2x - 1)
\][/tex]
3. Combine Factored Terms:
- Combine the factored forms within each group:
[tex]\[
6x^2(2x - 1) + 4(2x - 1)
\][/tex]
- Factor out the common binomial term [tex]\( (2x - 1) \)[/tex]:
[tex]\[
(2x - 1)(6x^2 + 4)
\][/tex]
### Verification:
1. Lucas' Factored Form: [tex]\((3x^2 + 2)(4x - 2)\)[/tex]
- Expand and verify if it matches the original polynomial:
[tex]\[
(3x^2 + 2)(4x - 2) = 3x^2 \cdot 4x + 3x^2 \cdot -2 + 2 \cdot 4x + 2 \cdot -2
\][/tex]
[tex]\[
= 12x^3 -6x^2 + 8x - 4
\][/tex]
- The factorization is correct as it gives the original polynomial.
2. Erick's Factored Form: [tex]\((2x - 1)(6x^2 + 4)\)[/tex]
- Expand and verify if it matches the original polynomial:
[tex]\[
(2x - 1)(6x^2 + 4) = 2x(6x^2) + 2x(4) - 1(6x^2) - 1(4)
\][/tex]
[tex]\[
= 12x^3 + 8x - 6x^2 - 4
\][/tex]
- The factorization is also correct as it gives the original polynomial.
### Conclusion:
Both Lucas and Erick have correctly grouped the terms of the polynomial [tex]\( 12x^3 - 6x^2 + 8x - 4\)[/tex] and factored it properly. Therefore, both students are correct in their approaches.
