How do you find the quotient of (x^4 - 5x^3 + x^2 - 6) div (x-1)(x4−5x3+x2−6)÷(x−1)?

1 Answer
Aaditya Raghavan
Sep 28, 2017

x^3-4x^2 - 3x - 3 - 9/(x-1)x3−4x2−3x−3−9x−1

Explanation:

x-1 ) x^4 - 5x^3 + x^2 - 6(x^3-4x^2-3x - 3x−1)x4−5x3+x2−6(x3−4x2−3x−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)

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