How do you find the quotient of #(x^4 - 5x^3 + x^2 - 6) div (x-1)#?

1 Answer
Sep 28, 2017

#x^3-4x^2 - 3x - 3 - 9/(x-1)#

Explanation:


#x-1 ) x^4 - 5x^3 + x^2 - 6(x^3-4x^2-3x - 3#

Each layer of division is listed below:

#I ... x^4 - 5x^3 - (x^4 - x^3) = x^4 - 5x^3 - x^4 + x^3 = -4x^3#

#II ... -4x^3 + x^2 -(-4x^3 + 4x^2) = -3x^2 #

#III ... -3x^2 - 6 - (-3x^2 + 3x) = -6 - 3x#

#IV ... -6 - 3x - (-3x + 3) = -9# (remainder)

To write the remainder, you put the remainder in the numerator of the fraction and the divisor in the denominator, at the end of the polynomial quotient:

#(x^4 - 5x^3 + x^2 - 6)/(x-1) = x^3-4x^2 - 3x - 3 - 9/(x-1)#