dd the polynomial expressions 2r2−3r+4

Question

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 \(10\).