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To factor the given polynomial \(2u^3 + 12u^2 - u - 6\) by grouping, follow these steps:

1. **Group terms**:
   \[
   (2u^3 + 12u^2) - (u + 6)
   \]

2. **Factor out the greatest common factor (GCF) from each group**:
   - From the first group \(2u^3 + 12u^2\), factor out \(2u^2\):
     \[
     2u^2(u + 6)
     \]
   - From the second group \(- (u + 6)\), factor out \(-1\):
     \[
     -1(u + 6)
     \]

3. **Factor out the common binomial factor \((u + 6)\)**:
   \[
   (2u^2 - 1)(u + 6)
   \]

4. **Check for further factorization**:
   - Notice that \(2u^2 - 1\) is a difference of squares:
     \[
     2u^2 - 1 = ( \sqrt{2}u - 1)(\sqrt{2}u + 1)
     \]
   - But if you stick to integer coefficients, then leave the factorization as:
     \[
     (2u^2 - 1)(u + 6)
     \]

Thus, the fully factored form of \(2u^3 + 12u^2 - u - 6\) is:

\[
(2u^2 - 1)(u + 6)
\]

Would you like more details or have any questions?

### Related Questions
1. How do you recognize terms that can be grouped for factoring?
2. What are common mistakes when factoring polynomials by grouping?
3. How can you factor a polynomial using synthetic division?
4. What is the difference of squares and how do you factor it?
5. How do you handle factoring polynomials with more than four terms?

### Tip:
When factoring by grouping, always check for a common factor among all terms before grouping, as it can simplify the problem significantly.

Factor the polynomial 2u^3 + 12u^2 - u - 6 by grouping.

Learn how to factor the polynomial 2u^3 + 12u^2 - u - 6 by grouping. This step-by-step solution includes factoring by grouping, identifying common factors, and applying the difference of squares theorem. Suitable for students in Grades 9-11.

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