Question

lucas and erick are factoring the polynomial 12x3 – 6x2 8x – 4. lucas groups the polynomial (12x3 8x) (–6x2 – 4) to factor. erick groups the polynomial (12x3 – 6x2) (8x – 4) to factor. who correctly grouped the terms to factor? explain.

Answer 1

Sample Response: Both students are correct because polynomials can be grouped in different ways to factor. Both ways result in a common binomial factor between the groups. Using the distributive property , this common binomial term can be factored out. Each grouping results in the same two binomial factors.

What did you include in your response? Check all that apply.

Polynomials can be grouped differently to factor.

Each way results in common binomials.

Each can be factored as a product of prime polynomials.

Explanation:

Answer 2

Both the students were correct.

Explanation:

Given : The polynomial is -
`12 x^3 - 6 x^2 + 8 x - 4`

To find : Who correctly grouped the terms to factor.

Lucas group the polynomial -
`( 12 x^3 + 8 x ) + ( - 6 x^2 - 4 )`

Now, we solve the Lucas polynomial

`(12x^3+8x) + (-6x^2-4)= 4x(3x^2+2 )-2(3x^2+2)`

`(12x^3+8x) + (-6x^2-4)= (3x^2+2 )(4x-2)`

`(12x^3+8x) + (-6x^2-4)= 2(3x^2+2)(2x-1)`

Erick group the polynomial -
`(12x^3-6x^2) + (8x- 4)`

Now, we solve the Erick polynomial

`(12x^3-6x^2) + (8x- 4)=6x^2(2x-1)+4(2x-1)`

`(12x^3-6x^2) + (8x- 4)=(6x^2+4)(2x-1)`

`(12x^3-6x^2) + (8x- 4)=2(3x^2+2)(2x-1)`

The factors of Lucas and Erick are same.

Each grouping leads to the same result.

Therefore, Both students are correct.