Question

find each sum.
$(3n^2 - 5n + 6) + (-8n^2 - 3n - 2) =$
\\(\

$$\begin{array}{l}\ \\ \\ \\ -11n^2 - 8n - 4\\\\\ \\ \\ \\ -11n^2 - 2n + 4\\\\\ \\ \\ \\ -5n^2 - 8n + 4\\\\\ \\ \\ \\ -5n^2 - 2n - 4\\\\\ \\end{array}$$

\\)
$(-4c^2 + 7cd + 8d) + (-3d + 8c^2 + 4cd) = \square c^2 + \square cd + \square d$

Explanation:

First Problem: \((3n^2 - 5n + 6) + (-8n^2 - 3n - 2)\)

Step1: Combine \(n^2\) terms

\(3n^2 + (-8n^2) = (3 - 8)n^2 = -5n^2\)

Step2: Combine \(n\) terms

\(-5n + (-3n) = (-5 - 3)n = -8n\)

Step3: Combine constant terms

\(6 + (-2) = 6 - 2 = 4\)

Step4: Combine all terms

Putting it together, we get \(-5n^2 - 8n + 4\)

Second Problem: \((-4c^2 + 7cd + 8d) + (-3d + 8c^2 + 4cd)\)

Step1: Combine \(c^2\) terms

\(-4c^2 + 8c^2 = (-4 + 8)c^2 = 4c^2\)

Step2: Combine \(cd\) terms

\(7cd + 4cd = (7 + 4)cd = 11cd\)

Step3: Combine \(d\) terms

\(8d + (-3d) = (8 - 3)d = 5d\)

Answer:

s:

1. For \((3n^2 - 5n + 6) + (-8n^2 - 3n - 2)\), the answer is \(\boldsymbol{-5n^2 - 8n + 4}\) (which matches the checked option).
2. For \((-4c^2 + 7cd + 8d) + (-3d + 8c^2 + 4cd)\), the blanks are filled as \(4\) (for \(c^2\)), \(11\) (for \(cd\)), and \(5\) (for \(d\)), so the expression is \(\boldsymbol{4c^2 + 11cd + 5d}\).