What is sum of coefficients of (x^9+2x^2-4x-1)/(x-1)?

Expert Answers
sciencesolve eNotes educator| Certified Educator

The problem provides the information that `x^9+2x^2-4x-1`  is divided by `x - 1` , hence, using the reminder theorem yields:

`x^9+2x^2-4x-1 = (x-1)(ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k) + m`

Notice that the degree of polynomial that represents the reminder is less than the degree of polynomial that represents the divisor, hence the reminder is the constant m.

You need to open the brackets such that:

`x^9+2x^2-4x-1 = ax^9 + bx^8 + cx^7 + dx^6 + ex^5 + fx^4 + gx^3 + hx^2 + kx - ax^8- bx^7 - cx^6- dx^5- ex^4- fx^3 -gx^2- hx- k + m`

Equating the coefficients of like terms yields:

`a = 1`

`b - a = 0 => b - 1 = 0 => b = 1`

`c - b = 0 => c - 1 = 0 => c = 1`

`d - c = 0 => d = 1`

`e - d = 0 => e = 1`

`f - e = 0 => f = 1`

`g - f = 0 => g = 1`

`h - g = 2 => h - 1 = 2 => h = 3`

`k - h = -4 => k - 3 = -4 => k = -1`

`-k + m = -1 => -1 + m = -1 => m = 0`

Since the reminder is 0, hence, the polynomial `x^9+2x^2-4x-1`  is divisible by `x - 1` .

You need to evaluate the sum of coefficients of quotient `ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k`  such that:

`a + b + c + d + e + f + g + h + k = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 3 - 1 = 9`

Hence, evaluating the sum of coefficients of quotient yields 9.