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Polynomial long division  How does it work? 
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#1
Jul1614, 04:04 AM

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Let's say I wanted to do the following calculation: (x^2 + 2x + 1) / (x+1)
I've scrolled through some online guides, and they all show how to do it, but not the principle behind it. I'm specifically having trouble with the fact, that instead of dividing the largest degree term with the entire denominator x+1 = x^{1}+x^{0}, they only take the largest degree term from the denominator, and divide it into the largest degree term in the numerator, as they progress down the steps. How can they just ignore the number 1 in this case. I know that once the largest degree terms have been divided, the result is then multiplied into the denominator and therefore the 1 is not completely ignored, but it just doesn't make sense to me. I don't see the connection between dividing 121 by 11, which is the same as dividing 121 = 1*10^2 + 2*10^1 + 1 * 10^0 by (1*10^1+1*10^0) which is equivalent to dividing x^2 + 2x + 1 by x+1. Could someone clarify the specifics for me? Why can I ignore the lower degree terms when doing the divisions? 


#2
Jul1614, 04:13 AM

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Each quotient term is the result of dividing the most significant dividend term by the most significant divisor term. The lower order terms in the dividend are not needed until they eventually become the most significant terms, and until then, zeroes can be used for the lower order terms in the dividend to calculate intermediate results.



#3
Jul1614, 04:53 AM

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It's not clear what you mean when you say that the lower degree terms are ignored when doing long division, be it with numbers or polynomials.
The long division algorithm is a stepped process: it generally takes more than one step to complete the process of finding the quotient. To take the example polynomial division: (x^2 + 2x + 1) / (x+1)



#4
Jul1614, 05:45 AM

P: 28

Polynomial long division  How does it work?
This is going to be awkward, since I don't know how to make a code window like SteamKing above, but if we were to replicate the method of polynomial long division with simple numerical long division, 121/11 = (100 + 20 + 1) / (10 + 1)
1) We would first divide 12 by 10 (from 11 = 10 + 1) and then multiply the result into 10 + 1 (the divisor), which would give us 1*10 + 1*1 = 10 + 1 = 11, the original divisor. 2) We would then subtract 10 + 1 from 12 = 10 + 2, which would give us 1. 3) Then we would take down the remaining less significant 1 from the dividend and add it to the 1 we got from the subtraction, resulting in 11. 4) Then we would divide 11 by 11 and get 1, multiply this one into the divisor again to get 11, subtract, and end up with the remainder 0. I do know what I'm supposed to do in long division, I'm just not comfortable with the 'why' of it. Why can I divide 12 by 10 and multiply the result into 11 to get the same answer as if I was dividing 12 by 11 and then multiplying the answer into 11? 12/11 and 12/10 do not produce the same remainder(the result of the subtraction procedure that is done during long division). 


#5
Jul1614, 06:54 AM

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#6
Jul1614, 08:33 AM

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To go along with no carries or borrows, you can also have negative terms in polynomial division whereas you don't allow that for "number" division. 


#7
Jul1614, 09:26 AM

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#8
Jul1614, 11:54 AM

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(just remember not to submit your reply if you're just looking at how he constructed his post, instead of replying to it) 


#9
Jul1614, 11:55 AM

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#10
Jul1614, 11:56 AM

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120 / 11 = 10, check 10 * 11 = 110 The next step subtracts 110 from 121, leaving 11. The quotient of 11/11 = 1. Adding the two partial results together produces Q = 10 + 1 = 11, which is 121/11. 


#11
Jul1614, 03:15 PM

P: 65

Polynomial long division can be viewed as a more general case of the Euclidean algorithm. The basic procedure is similar to integers. At each step ##k##, a quotient polynomial ##q_k(x)## and a remainder polynomial ##r_k(x)## are identified to satisfy the recursive equation
$$r_{k−2}(x) = q_k(x) r_{k−1}(x) + r_k(x),$$ where ##r_{−2}(x) = a(x)## and ##r_{−1}(x) = b(x)##. The quotient polynomial is chosen so that the leading term of ##q_k(x) r_{k−1}(x)## equals the leading term of ##r_{k−2}(x)##; this ensures that the degree of each remainder is smaller than the degree of its predecessor ##\deg[r_k(x)] \lt \deg[r_{k−1}(x)]##. Since the degree is a nonnegative integer, and since it decreases with every step, the Euclidean algorithm concludes in a finite number of steps. The final nonzero remainder is the greatest common divisor of the original two polynomials, ##a(x)## and ##b(x)##. You can gain a deeper understanding by viewing this as a special case of multivariate polynomial division algorithms, such as the Gröbner basis algorithm. One gains further insight from this more general perspective of monomial orderings. 


#12
Jul1714, 01:32 AM

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#13
Jul1714, 10:43 AM

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My companion have said right that each remainder term is the result of dividing the most significant divided term by the most significant divisor term. The lower terms in the divided are not required until they inevitably turned into the most significant terms, and until then, zeroes could be utilized for the lower request terms in the profit to compute transitional results.



#14
Jul1714, 11:23 AM

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#15
Jul1714, 03:02 PM

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Some students do not really understand long division algorithm in grade school  instead, they really understand well enough a few years AFTER grade school, and only at this time the algorithm really makes sense. Polynomial long division as taught in ninthgrade Algebra 1 is one of those things that pushes a student into real progress for ordinary long division. Suddenly, long division seems to make sense. 


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