Long Division with Polynomials

Apr 2016
58
3
Australia
Hi everyone!

I am VERY confused on how to do long division with polynomials!!
Here is an example:

3x^3- 4x^2+ 2x +1, x+2

Any help would be much appreciated!
Thank you in advance :)

~EbonyJade :)
 
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Feb 2015
227
15
california
\(\displaystyle \frac{3x^2-4x^2+2x+1}{x+2} \).
Lets look at numeric division first. I'll ask you how you divide 906 by 3 by hand. Here is the usual way: You would look at 906 digit by digit. You would ask yourself, "What times 3 is 9 (the first digit in 906). 3 times 3 is nine. You know that 3 would be the first digit of your quotient. Then you would continue and say 3*0 = 0, and 3*1 =3. Thus your quotient is 301

The trick in polynomial division is to look at the dividend \(\displaystyle (3x^2-4x^2+2x+1)\) one term at a time. So first you ask yourself what can you multiply by x+2 to get a factor of 3x^2. Well 3x times x is 3x^2. This matches the first term of your dividend. so \(\displaystyle 3x^2 * (x+2) = 3x^3 + 2x^2 \). Now you subtract \(\displaystyle 3x^3+2x^2 \) from \(\displaystyle 3x^3 -4x^2 + 2x + 1\) You would get \(\displaystyle -6x^2+2x+1 \) Now you see what times x+2 equals the first term of \(\displaystyle -6x^2+2x+1 \). You can see that \(\displaystyle -6(x+2) = -6x^2 - 6x \). Good. Your first terms match. Execute the following subtraction: \(\displaystyle (-6x^2+2x+1) - ( -6x^2 - 6x) \) to get \(\displaystyle 8x + 1 \).... Continue doing this procedure until you can't multiply (x+2) by \(\displaystyle Cx^n \) where C is a real number, and n is a positive real number. The remaining term would be your remainder.
 
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romsek

MHF Helper
Nov 2013
6,746
3,037
California
what toesockshoe said in picture form

Clipboard01.jpg
 
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Apr 2016
58
3
Australia
Wow! Thankyou both very much!! That helps a lot! :) :)