Divide 3m^5+2n^3+m+1 by m^2+1

Answer 1

To divide the polynomial \(3m^5 + 2n^3 + m + 1\) by \(m^2 + 1\), we will use polynomial long division.

1. Setup the long division: We will divide \(3m^5\) by \(m^2\).

   • \( \frac{3m^5}{m^2} = 3m^3\)

2. Multiply and subtract: Multiply \(3m^3\) by \(m^2 + 1\):

   \[ 3m^3(m^2 + 1) = 3m^5 + 3m^3 \]

   Now subtract this from the original polynomial:

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

3. Continue dividing: Now we divide \(-3m^3\) by \(m^2\):

   • \( \frac{-3m^3}{m^2} = -3m\)

4. Multiply and subtract again: Multiply \(-3m\) by \(m^2 + 1\):

   \[ -3m(m^2 + 1) = -3m^3 - 3m \]

   Now subtract this from the current polynomial:

   \[ (2n^3 - 3m^3 + m + 1) - (-3m^3 - 3m) = 2n^3 + 4m + 1 \]

5. Final division step: Now take \(2n^3\) (the leading term) and divide it by \(m^2\). Since \(n\) is not related to \(m\), we cannot divide it in terms of \(m\), so we can't get any more \(m\) terms.

Thus, the remainder is \(2n^3 + 4m + 1\).

Final Result

The result of dividing \(3m^5 + 2n^3 + m + 1\) by \(m^2 + 1\) is:

\[ 3m^3 - 3m + \frac{2n^3 + 4m + 1}{m^2 + 1} \]

Or more succinctly:

\[ 3m^3 - 3m + \frac{2n^3 + 4m + 1}{m^2 + 1} \]