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Posted by **Sarah** on Thursday, September 15, 2011 at 5:18pm.

#9. 3x^4 – 2x^3 + 5x^2 + x + 1;x^2 + 2x

- Discrete Math -
**MathMate**, Thursday, September 15, 2011 at 5:39pmDo exactly as you would do in a long division of numbers:

3x^4 – 2x^3 + 5x^2 + x + 1;x^2 + 2x

I will do it below, but the coefficients will not be lined up because I do not know how to insert spaces that don't get merged at this forum.

Divide x^2+2x into

3x^4 – 2x^3 + 5x^2 + x + 1

First divide 3x^4 by x^2 to get**3x^2**.

Then multiply (x^2+2x) by 3x^2 to get

3x^4+6x^3

Subtract 3x^4+6x^3 from 3x^4 – 2x^3 + 5x^2 + x + 1 to get

– 8x^3 + 5x^2 + x + 1

Repeat the same as above:

Divide -8x^3 by x^2 to get**-8x**

Multiply x^2+2x by -8x to get -8x^3 -16x^2.

Subtract -8x^3 -16x^2 from – 8x^3 + 5x^2 + x + 1 to get 21x^2 +x + 1.

The same process will repeat itself to get the final answer by collecting all the above quotient terms:

3x^2-8x+21 with a remainder of -41x+1

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