Which Shows One Way To Determine The Factors Of $12x^3 - 2x^2 + 18x - 3$ By Grouping?A. $2x^2(6x - 1) + 3(6x - 1$\]B. $2x^2(6x - 1) - 3(6x - 1$\]C. $6x(2x^2 - 3) - 1(2x^2 - 3$\]D. $6x(2x^2 + 3) + 1(2x^2 + 3$\]

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Introduction

In algebra, factoring polynomials is an essential skill that helps us simplify complex expressions and solve equations. One method of factoring polynomials is by grouping, which involves rearranging the terms of a polynomial to facilitate factoring. In this article, we will explore how to determine the factors of a given polynomial by grouping.

Understanding the Concept of Grouping

Grouping is a factoring technique that involves rearranging the terms of a polynomial to create two or more groups of terms that can be factored separately. This method is particularly useful when the polynomial has multiple terms with common factors. By grouping the terms, we can identify the common factors and factor them out, making it easier to simplify the polynomial.

The Given Polynomial

The given polynomial is $12x^3 - 2x^2 + 18x - 3$. Our goal is to determine the factors of this polynomial by grouping.

Option A: 2x2(6x−1)+3(6x−1)2x^2(6x - 1) + 3(6x - 1)

Let's examine Option A: 2x2(6x−1)+3(6x−1)2x^2(6x - 1) + 3(6x - 1). To determine if this is the correct grouping, we need to check if the terms can be factored out.

$2x^2(6x - 1) + 3(6x - 1)$
= (2x^2 + 3)(6x - 1)

As we can see, the terms can be factored out, and we have a common binomial factor (6x−1)(6x - 1).

Option B: 2x2(6x−1)−3(6x−1)2x^2(6x - 1) - 3(6x - 1)

Now, let's examine Option B: 2x2(6x−1)−3(6x−1)2x^2(6x - 1) - 3(6x - 1). To determine if this is the correct grouping, we need to check if the terms can be factored out.

$2x^2(6x - 1) - 3(6x - 1)$
= (2x^2 - 3)(6x - 1)

As we can see, the terms can be factored out, and we have a common binomial factor (6x−1)(6x - 1).

Option C: 6x(2x2−3)−1(2x2−3)6x(2x^2 - 3) - 1(2x^2 - 3)

Next, let's examine Option C: 6x(2x2−3)−1(2x2−3)6x(2x^2 - 3) - 1(2x^2 - 3). To determine if this is the correct grouping, we need to check if the terms can be factored out.

$6x(2x^2 - 3) - 1(2x^2 - 3)$
= (6x - 1)(2x^2 - 3)

As we can see, the terms can be factored out, and we have a common binomial factor (2x2−3)(2x^2 - 3).

Option D: 6x(2x2+3)+1(2x2+3)6x(2x^2 + 3) + 1(2x^2 + 3)

Finally, let's examine Option D: 6x(2x2+3)+1(2x2+3)6x(2x^2 + 3) + 1(2x^2 + 3). To determine if this is the correct grouping, we need to check if the terms can be factored out.

$6x(2x^2 + 3) + 1(2x^2 + 3)$
= (6x + 1)(2x^2 + 3)

As we can see, the terms can be factored out, and we have a common binomial factor (2x2+3)(2x^2 + 3).

Conclusion

In conclusion, we have examined four different options for grouping the given polynomial $12x^3 - 2x^2 + 18x - 3$. We have found that all four options can be factored out, but only one option has a common binomial factor that can be factored out.

The correct answer is Option B: 2x2(6x−1)−3(6x−1)2x^2(6x - 1) - 3(6x - 1), which can be factored out as (2x2−3)(6x−1)(2x^2 - 3)(6x - 1).

Final Answer

Q: What is the concept of grouping in algebra?

A: Grouping is a factoring technique that involves rearranging the terms of a polynomial to create two or more groups of terms that can be factored separately. This method is particularly useful when the polynomial has multiple terms with common factors.

Q: How do I determine the factors of a polynomial by grouping?

A: To determine the factors of a polynomial by grouping, follow these steps:

  1. Rearrange the terms of the polynomial to create two or more groups of terms.
  2. Check if the terms in each group have a common factor.
  3. If the terms have a common factor, factor it out.
  4. Repeat the process until all the terms have been factored out.

Q: What are some common mistakes to avoid when factoring by grouping?

A: Some common mistakes to avoid when factoring by grouping include:

  • Not rearranging the terms correctly to create groups of terms with common factors.
  • Not checking if the terms in each group have a common factor.
  • Factoring out a term that is not a common factor of all the terms in the group.

Q: Can I use grouping to factor polynomials with multiple variables?

A: Yes, you can use grouping to factor polynomials with multiple variables. However, you need to be careful when rearranging the terms to create groups of terms with common factors.

Q: How do I know if a polynomial can be factored by grouping?

A: A polynomial can be factored by grouping if it has multiple terms with common factors. You can check if a polynomial can be factored by grouping by rearranging the terms and looking for common factors.

Q: What are some examples of polynomials that can be factored by grouping?

A: Some examples of polynomials that can be factored by grouping include:

  • 12x3−2x2+18x−312x^3 - 2x^2 + 18x - 3
  • 2x2(6x−1)+3(6x−1)2x^2(6x - 1) + 3(6x - 1)
  • 6x(2x2−3)−1(2x2−3)6x(2x^2 - 3) - 1(2x^2 - 3)

Q: What are some examples of polynomials that cannot be factored by grouping?

A: Some examples of polynomials that cannot be factored by grouping include:

  • x2+2x+1x^2 + 2x + 1
  • 2x2+5x+32x^2 + 5x + 3

Q: Can I use grouping to factor polynomials with negative coefficients?

A: Yes, you can use grouping to factor polynomials with negative coefficients. However, you need to be careful when rearranging the terms to create groups of terms with common factors.

Q: How do I check if a polynomial has been factored correctly by grouping?

A: To check if a polynomial has been factored correctly by grouping, follow these steps:

  1. Multiply the factors together to get the original polynomial.
  2. Check if the resulting polynomial is the same as the original polynomial.
  3. If the resulting polynomial is the same as the original polynomial, then the polynomial has been factored correctly by grouping.

Conclusion

In conclusion, factoring polynomials by grouping is a useful technique that can be used to simplify complex expressions and solve equations. By following the steps outlined in this article, you can determine the factors of a polynomial by grouping and check if a polynomial has been factored correctly.