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(9c² + 7c + 8) - (7c² + 4c + 1)

Question

(9c² + 7c + 8) - (7c² + 4c + 1)

Explanation:

Step1: Distribute the negative sign

We need to distribute the negative sign to each term inside the second parentheses:
\((9c^{2}+7c + 8)-(7c^{2}+4c + 1)=9c^{2}+7c + 8-7c^{2}-4c - 1\)

Step2: Combine like terms

  • For the \(c^{2}\) terms: \(9c^{2}-7c^{2}=(9 - 7)c^{2}=2c^{2}\)
  • For the \(c\) terms: \(7c-4c=(7 - 4)c = 3c\)
  • For the constant terms: \(8-1 = 7\)

Combining these results, we get \(2c^{2}+3c + 7\)

Answer:

\(2c^{2}+3c + 7\)