Factor completely.

5n5–n4+5n3–n2

1 answer

To factor the expression \( 5n^5 - n^4 + 5n^3 - n^2 \), we can first group the terms in a way that makes factoring easier. Let's rearrange the expression:

\[ 5n^5 + 5n^3 - n^4 - n^2. \]

Next, we can group the terms:

\[ (5n^5 + 5n^3) + (-n^4 - n^2). \]

Now we can factor out common factors from each group:

  1. From the first group \( 5n^5 + 5n^3 \), we can factor out \( 5n^3 \):

\[ 5n^3(n^2 + 1). \]

  1. From the second group \( -n^4 - n^2 \), we can factor out \(-n^2\):

\[ -n^2(n^2 + 1). \]

Now, our expression becomes:

\[ 5n^3(n^2 + 1) - n^2(n^2 + 1). \]

Notice that we have a common factor of \((n^2 + 1)\):

\[ (5n^3 - n^2)(n^2 + 1). \]

Next, we can factor out \(n^2\) from the first part:

\[ n^2(5n - 1)(n^2 + 1). \]

Putting it all together, the completely factored form of the expression \( 5n^5 - n^4 + 5n^3 - n^2 \) is:

\[ n^2(5n - 1)(n^2 + 1). \]