How do you divide (3x^4 + 2x^3 - 11x^2 - 2x + 5)/(x^2 - 2)3x4+2x3−11x2−2x+5x2−2?

1 Answer
George C.
Nov 7, 2015

Use long division of the coefficients to find:

3x^4+2x^3-11x^2-2x+5 = (x^2-2)(3x^2+2x-5) + 2x-53x4+2x3−11x2−2x+5=(x2−2)(3x2+2x−5)+2x−5

That is:

(3x^4+2x^3-11x^2-2x+5)/(x^2-2) = 3x^2+2x-5 + (2x-5)/(x^2-2)3x4+2x3−11x2−2x+5x2−2=3x2+2x−5+2x−5x2−2

Explanation:

Long divide the coefficients, using a long division similar to division of integers:

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Note that the divisor is 1, 0, -21,0,−2 to represent x^2+0x-2x2+0x−2 including the term in xx.

Choose the first term 33 of the quotient to match the leading term of the dividend when multiplied by the divisor.

Multiply the divisor by 33 to get 3, 0, -63,0,−6 and subtract from the dividend to get a remainder. Bring down the next term from the dividend alongside it and then repeat to get the next term of the quotient, etc. Stop when the degree of the running remainder is less than the divisor.

3x^4+2x^3-11x^2-2x+5 = (x^2-2)(3x^2+2x-5) + 2x-53x4+2x3−11x2−2x+5=(x2−2)(3x2+2x−5)+2x−5

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