To factor the expression $2n^3 - 27n^2 + 81n$, let's follow these steps:

Step 1: Factor out the greatest common factor (GCF)

First, identify the GCF of the terms. Here, each term contains a factor of $n$, and the greatest common factor of the coefficients $2$, $-27$, and $81$ is $1$, as there is no common factor greater than 1 for the coefficients.

Thus, we can factor out an $n$:

$2n^3 - 27n^2 + 81n = n(2n^2 - 27n + 81)$

Step 2: Factor the quadratic expression

Now, we need to factor the quadratic expression $2n^2 - 27n + 81$.

We can factor this quadratic using the method of factoring by grouping or finding two numbers that multiply to $2 \times 81 = 162$ and add to $-27$.

The two numbers that satisfy this are $-9$ and $-18$, because: $-9 \times -18 = 162 \quad \text{and} \quad -9 + (-18) = -27$

Now, rewrite the middle term $-27n$ as $-9n - 18n$:

$2n^2 - 9n - 18n + 81$

Group the terms:

$(2n^2 - 9n) - (18n - 81)$

Factor out the common factors from each group:

$n(2n - 9) - 9(2n - 9)$

Now factor out the common binomial factor $(2n - 9)$:

$(2n - 9)(n - 9)$

Step 3: Write the final factorization

So, the fully factored form of the expression is:

$2n^3 - 27n^2 + 81n = n(2n - 9)(n - 9)$

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Would you like a more detailed explanation or have any questions on this?

Here are five related questions:

1. How do you factor expressions with a common variable factor?
2. What is the method for factoring a quadratic trinomial when the leading coefficient is not 1?
3. How do you check if your factorization is correct?
4. Can all cubic expressions be factored easily?
5. How would you factor a cubic polynomial if the GCF isn't obvious?

Tip: Always check if there's a GCF first, as it can simplify the factorization process significantly.