Question

dd the polynomial expressions 2r2−3r+4
and −2r2+3r+6
.

Answer 1

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

1. Start with the expressions:
\[
(2r^2 - 3r + 4) + (-2r^2 + 3r + 6)
\]

2. Group the like terms:
\[
(2r^2 - 2r^2) + (-3r + 3r) + (4 + 6)
\]

3. Simplify each group:
- For the \(r^2\) terms: \(2r^2 - 2r^2 = 0\)
- For the \(r\) terms: \(-3r + 3r = 0\)
- For the constant terms: \(4 + 6 = 10\)

4. Combine the results:
\[
0 + 0 + 10 = 10
\]

Thus, the sum of the polynomials \(2r^2 - 3r + 4\) and \(-2r^2 + 3r + 6\) is \(10\).