Question

Add the polynomial expressions $2r^{2}-3r+4$ and $-2r^{2}+3r+6$ (1 point)
$\square $

Answer 1

To add the polynomial expressions \(2r^2 - 3r + 4\) and \(-2r^2 + 3r + 6\), we combine like terms.

1. Identify the coefficients of corresponding terms:
- For \(r^2\): \(2r^2\) and \(-2r^2\)
- For \(r\): \(-3r\) and \(3r\)
- For the constant terms: \(4\) and \(6\)

2. Add the coefficients:
- \(r^2\) terms: \(2r^2 - 2r^2 = 0r^2\)
- \(r\) terms: \(-3r + 3r = 0r\)
- Constant terms: \(4 + 6 = 10\)

Putting it all together, we have:
\[
0r^2 + 0r + 10 = 10
\]

Thus, the final result of adding the two polynomial expressions is:
\[
\boxed{10}
\]