Question

a, b and c are polynomials, where a = n, b = 2n + 6, and c = n² - 1. what is ab - c in simplest form? 3n² + 5 -n² + 3n + 5 2n² + 6n - 1 n² + 6n + 1

Explanation:

Step1: Calculate \( AB \)

Given \( A = n \) and \( B = 2n + 6 \), multiply them:
\( AB = n(2n + 6) = 2n^2 + 6n \)

Step2: Subtract \( C \) from \( AB \)

Given \( C = n^2 - 1 \), compute \( AB - C \):
\( AB - C = (2n^2 + 6n) - (n^2 - 1) \)
Distribute the negative sign: \( 2n^2 + 6n - n^2 + 1 \)

Step3: Combine like terms

Combine \( 2n^2 - n^2 = n^2 \), and the remaining terms are \( 6n + 1 \)? Wait, no—wait, recheck:
Wait, \( (2n^2 + 6n) - (n^2 - 1) = 2n^2 + 6n - n^2 + 1 = (2n^2 - n^2) + 6n + 1 = n^2 + 6n + 1 \)? Wait, no, wait the options have \( n^2 + 6n + 1 \) as the last option. Wait, did I make a mistake? Wait, let's re-express:

Wait, \( AB = n(2n + 6) = 2n^2 + 6n \). Then \( AB - C = 2n^2 + 6n - (n^2 - 1) = 2n^2 + 6n - n^2 + 1 = n^2 + 6n + 1 \). Yes, that's the last option.

Answer:

\( n^2 + 6n + 1 \) (corresponding to the option: \( n^2 + 6n + 1 \))