Divide (2x^4-x^2+6)/(x^2+3x+2)

Problem

(2*x^4−x^2+6)/(x^2+3*x+2)

Solution

1. Set up the polynomial long division by writing the dividend 2*x^4+0*x^3−x^2+0*x+6 and the divisor x^2+3*x+2

2. Divide the leading term of the dividend 2*x^4 by the leading term of the divisor x^2 to get the first term of the quotient, 2*x^2

3. Multiply 2*x^2 by the divisor x^2+3*x+2 to get 2*x^4+6*x^3+4*x^2 then subtract this from the dividend to get the new remainder −6*x^3−5*x^2+0*x+6

4. Divide the leading term of the new remainder −6*x^3 by x^2 to get the next term of the quotient, −6*x

5. Multiply −6*x by the divisor to get −6*x^3−18*x^2−12*x then subtract this from the current remainder to get 13*x^2+12*x+6

6. Divide the leading term 13*x^2 by x^2 to get the final constant term of the quotient, 13

7. Multiply 13 by the divisor to get 13*x^2+39*x+26 then subtract this from the current remainder to find the final remainder −27*x−20

8. Express the result as the sum of the quotient and the remainder over the divisor.

Final Answer

(2*x^4−x^2+6)/(x^2+3*x+2)=2*x^2−6*x+13+(−27*x−20)/(x^2+3*x+2)

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