Divide (6x^3-16x^2+17x-6)/(3x-2)

Problem

(6*x3−16*x2+17*x−6)/(3*x−2)

Solution

1. Divide the first term of the numerator 6*x3 by the first term of the divisor 3*x

(6*x3)/(3*x)=2*x2

2. Multiply the entire divisor 3*x−2 by 2*x2 and subtract the result from the original polynomial.

(6*x3−16*x2+17*x−6)−(6*x3−4*x2)=−12*x2+17*x−6

3. Divide the first term of the new polynomial −12*x2 by the first term of the divisor 3*x

(−12*x2)/(3*x)=−4*x

4. Multiply the divisor 3*x−2 by −4*x and subtract the result from the current polynomial.

(−12*x2+17*x−6)−(−12*x2+8*x)=9*x−6

5. Divide the first term of the remaining polynomial 9*x by the first term of the divisor 3*x

(9*x)/(3*x)=3

6. Multiply the divisor 3*x−2 by 3 and subtract to find the remainder.

(9*x−6)−(9*x−6)=0

Final Answer

(6*x3−16*x2+17*x−6)/(3*x−2)=2*x2−4*x+3

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