The function \(y=\frac{x^3+8x^2+21x+18}{x+2}\) can be simplified into the function \(y=Ax^2+Bx+C\), defined everywhere except at x = D. What is the sum of the values of A, B, C, and D?

I got the answer of 13, but it said incorrect. I'm thinking maybe the answer is 14, on further consideration.

Thanks!

PurpleWasp Aug 8, 2024

#1**0 **

(x ^ 3 + 8x ^ 2 + 21x + 18) can be factored into ( x + 2 )( x + 3 )( x + 3 ). We divide by ( x + 2 ), and they cancel. Expanding

( x + 3 ) ^ 2 gives us x ^ 2 + 6x + 9. A = 1, B = 6, and C = 9. D is the one value that x can't be. If x was -2, then we would be dividing by 0, which absolutely can't happen. So, D = -2. Adding them together, A + B + C + D = 1 + 6 + 9 + (-2) = **14**.

PurpleWasp Aug 9, 2024