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I don't understand polynomial division

  1. Oct 26, 2014 #1


    At first he shows 2x+4 / 2 and you just divide both 2x and 4 by 2. But then in the next example he is dividing x^2+3x+6 by x+1 and he doesn't divide x^2 by x+1, 3x by x+1 and 6 by x+1. I do not understand how he does the problem.
     
  2. jcsd
  3. Oct 26, 2014 #2

    Mark44

    Staff: Mentor

    The first step is to divide x2 by x, resulting in a partial quotient of x. To see how close this is, multiply that result by x + 1, to get x2 + x.

    Next, you subtract x2 + x from x2 + 3x, which results in 2x.

    Bring down the next term, 6, and tack it onto the 2x, resulting in 2x + 6.

    Divide 2x + 6 by x, which results in 2. See how close you got by multiplying 2 times x + 1 to get 2x + 2.

    Subtract 2x + 2 from 2x + 6 to get 4, which is the final remainder. Since the degree of 4 is zero and the degree of x + 1 is 1, you stop.

    Polynomial division is sort of like ordinary division of numbers. Take as an example 155 divided by 12.

    To make this look a bit more like polynomial division, write 12 as 10 + 2, and write 155 as 100 + 50 + 5.
    First, divide 100 by 10, which is 10. Write this 10 above the 50. Check how close this was by multiplying 10 times (10 + 2) which is 100 + 20.

    Subtract 100 + 20 from the line above, which is 100 + 50. The difference is 30.

    Bring down the final number, 5.

    Divide 30 + 5 by 10, to get 3. To check how close we are, multiply 10 + 2 by 3 to get 30 + 6. Since this is larger than the number on the line above (30 + 5), we need to revise our partial answer downward. Trying again, we get a new partial answer of 2. Check by multiplying 10 + 2 by 2 to get 20 + 4. Subtract 20 + 4 from 30 + 5, which is 10 + 1 (or 11). Since 10 + 1 is smaller than our divisor, we're done.

    We have shown that 155 divided by 12 is 12, with a remainder of 11.
     
  4. Oct 26, 2014 #3

    BvU

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    Compare with the long division of 136 by 11: you don't start with 100/11, continue with 30/11 and end with 6/11, right ?

    (So I just grabbed an easy value for x, namely 10).

    What you do is you determine the coefficient (call it a) for the highest power of x in the quotient: (x2+3x+6 )/(x+1) = (ax + ...) and you subtract what you have so far from the dividend :

    (x2+3x+6 )/(x+1) = ax + (x2+3x+6 ) - ax * (x+1) ) / (x+1)

    Of course here a = 1, so you continue with (x2+3x+6 )/(x+1) = ax + (2x+6 ) / (x+1)


    Well well, two simultaneous responses!
     
  5. Oct 26, 2014 #4

    SteamKing

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    In the first example, you can factor 2x + 4 into 2 * (x+2). If you then divide 2 * (x + 2) by 2, the twos cancel, leaving x+2 as the quotient. This is similar to dividing 75 by 5: you can write 75 as 15 * 5, which when divided by 5, leaves 15 as the quotient.

    Now, dividing x2 + 3x + 6 by x + 1 is not as simple, because the factors of x2 + 3x + 6 are not immediately obvious. Here, one must fall back on the long division algorithm, where you find out how many times the divisor, x + 1, will go into the dividend, x2 + 3x + 6. In this case, x + 1 goes into the dividend x times:

    Code (Text):

                          x
                        _______________
                   x+1  ) x^2 + 3x + 6
                          x^2 + x
                        ---------------
                               2x + 6
     
    The first part of the quotient is x with 2x+6 left over. Since this remainder is of at least equal degree as the divisor, a further partial division can be carried out:

    Code (Text):

                          x+2
                        _______________
                   x+1  ) x^2 + 3x + 6
                          x^2 +  x
                        ---------------
                                2x + 6
                                2x + 2
                        ---------------
                                     4
     
    So, x2 + 3x + 6 divided by x+1 has a quotient of x+2 with remainder 4. You can always check the division by multiplying the quotient times the divisor and adding any remainder to the result. The answer should be the original dividend.

    Perhaps you should practice first by doing long division using numbers, and then try it with polynomials.
     
  6. Oct 28, 2014 #5
    Why? Isn't x+1 supposed to go in?

    What do you mean "how close" it is? How close it is to what?
    No, you divide it into 13 first and carry 2 over onto the 6. I didn't get the rest of your post.

    What do you mean it goes into the dividen x times? In the previous example you never mentioned a dividend.
     
  7. Oct 28, 2014 #6

    Mark44

    Staff: Mentor

    Assuming that you were replying to my post, actually I do start with 100/11, then 30/11, and finally 6/11, but only to draw a parallel to polynomial division.
     
  8. Oct 28, 2014 #7

    RUber

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    You can't divide ##x^2## by ##x+1## evenly, so you are only comparing the highest order terms at any one time.
    You know that ##(x+1)*x = x^2+x##, so you put that in as your best guess. In any case, you wont have any ##x^2## terms left over.
    Then you move on to whatever is left over.
    You can repeat this process as many times as you want, but most people finish with the ##\frac 1{x+1}## term, or the ##x^{-1}## place.
    The goal at each step is to eliminate the highest order term in your numerator.
     
  9. Oct 28, 2014 #8
    I don't get it... What do you mean comparing the highest order terms at any one time?
    Best guess at what? How many times x+1 goes in? Like you said, it doesn't go in x^2+x times.
    Whatever is left over of what where?
    This is so confusing and I don't get it.
     
  10. Oct 28, 2014 #9

    SteamKing

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    I'm sorry. I thought these terms were clear, or could at least be looked up.

    For an expression like a = b/c, b is called the dividend, c is called the divisor, and a is called the quotient. These terms are the same regardless of whether one is dividing numbers or polynomials.

    Like I said. If you want to understand polynomial division, practice using the long division algorithm on numbers first.
     
  11. Oct 28, 2014 #10

    Mark44

    Staff: Mentor

    It's much easier to figure out how many times x goes into x2, isn't it? And x is reasonably close to x + 1.
    We're actually dividing x2 + 3x (the first two terms of the dividend, the thing that we're dividing) by x + 1 (the divisor), and we got a partial answer of x. To see how close this answer is to the correct answer, we multiply our partial quotient (x) by the divisor (x + 1), and get x2 + x. This isn't quite the same as x2 + 3x, so we subtract x2 + x from x2 + 3x to see how close our answer was. If it was spot on, the two quantities would have been the same and we would have gotten 0 instead of 2x that you see below.
    Code (Text):

                                 x
                        _______________
                   x+1  ) x^2 + 3x + 6
                          x^2 + x
                        ---------------
                               2x
     
    Since our remainder, 2x, is the same degree as the divisor (x + 1) - both are degree 1, we're not done yet.

    We continue by bringing down the next term from the dividend (+6)
    Code (Text):

                                 x
                        _______________
                   x+1  ) x^2 + 3x + 6
                          x^2 + x
                        ---------------
                               2x + 6
     
    Now we divide 2x in the bottom line by x in the divisor, and get 2, and write that at the top (in the quotient).
    Code (Text):

                                 x + 2
                        _______________
                   x+1  ) x^2 + 3x + 6
                          x^2 + x
                        ---------------
                               2x + 6
     
    Just as before, we see how close our partial answer is to the actual answer by multiplying the 2 in the quotient by the divisor (x + 1), to get 2x + 2.
    Code (Text):

                                 x + 2
                        _______________
                   x+1  ) x^2 + 3x + 6
                          x^2 + x
                        ---------------
                               2x + 6
                               2x + 2
                              ---------
                                    4
     
    How close did we come? We should have gotten exactly 2x + 6, but we instead got 2x + 2. If we subtract again (which is always how you check how close two things are) we get a final remainder of 4. Since the degree of 4 is 0 (there are no x terms) and the degree of the divisor is 1, we are done.

    What this shows is that x2 + 3x + 6 = (x + 1)(x + 2) + 4. Since there is a remainder of 4, this means that x + 1 does not divide x2 + 3x + 6 evenly.
    It also shows that
    $$\frac{x^2 + 3x + 6}{x + 1} = x + 2 + \frac{4}{x + 1}$$
     
  12. Oct 28, 2014 #11

    Mark44

    Staff: Mentor

    Yes.
     
  13. Oct 28, 2014 #12

    RUber

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    Think about when you learned how to FOIL or factor simple polynomials:
    ##(x+2)(x+4)=x^2+6x+8##
    To go in reverse and factor, you know that the leading terms are probably x, since ##x*x=x^2##.
    (x+a) (x+b), then you look for a and b such that ab=8 and a+b=6.
    If you were given one of the factors, say, (x+2), you might be able to just figure out what the other one was.
    You could also do some long division.
    ##\frac{x^2+6x+8}{x+2} =\frac{x^2+6x}{x+2}+\frac{8}{x+2}##
    ##=x\frac{x+6}{x+2}+\frac{8}{x+2}=x\left(\frac{x+2}{x+2}+\frac{4}{x+2}\right)+\frac{8}{x+2}##
    ##=x\left(1+\frac{4}{x+2}\right)+\frac{8}{x+2}=x+\frac{4x+8}{x+2}=x+4##
    The long division framework is a more efficient way to keep the algebra clear.
     
  14. Oct 28, 2014 #13
    Yeah I still don't understand your explanation.
     
  15. Oct 28, 2014 #14
    What do you mean reasonably close? If I divide 12 by 6 I don't divide 12 by 5 because 5 is reasonably close to 6.
    How is it a 'partial answer'? I don't get it.
     
  16. Oct 28, 2014 #15
    I don't get how talking about factoring led to the long division way.
     
  17. Oct 28, 2014 #16

    Mark44

    Staff: Mentor

    But you could write 6 as 5 + 1 and 12 as 10 + 2. Then this division problem would look more like the polynomial division we're trying to explain here.
    Code (Text):

                               2
                        ________
                   5+1  ) 10 + 2
                          10 + 2
                        ---------
                               0
     
    First, divide 10 by 5, getting 2, which I place at the top. At this point, 2 is the partial quotients.,
    Next, multiply 2 times (5 + 1), which gives me 10 + 2. Write that below the dividend (the thing that's being divided.
    Subtract: 10 + 2 - (10 - 2), which is 0.
    We're done.
    x is a partial answer because we're not done yet - we don't have the complete answer. Look at the sequence of steps that I laid out in the code blocks in post #10, and reread the explanations that I give with them. The first partial quotient was just x. We keep refining the process, getting more terms in the quotient until we're done, with a final quotient of x + 2, plus a remainder.
     
  18. Oct 31, 2014 #17
    I was taught how to do long division in school but I don't understand it even after watching 5-10 videos and reading about it now.

    Like in this video:


    You put 8 in the 10s place. That implies it's 80 yes, but it's not 80 as we can see when it's finished. So why leave out the 6? It's 86, not 8, not 80. Also, in the middle of it, he multiplies 8 by 4 and gets 32 and brings it under the 34. Why?
     
  19. Oct 31, 2014 #18

    RUber

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    Yes, 8 in the 10s place is 80. 6 in the ones place is 6. 80+6 is 86.

    Long division is all about sequential approximation.

    for example if you have: 537/32
    Starting with 10s, 20x32 is too much, 10x32 is a good starting point. (If you wanted to start with 12 or 15 it wouldn't hurt the process, but in long division you normally do one place column at a time. )
    537 = 320 + 217.
    So I know that 537/32 = 10 + 217/32.
    I also know that 217/32 is less than 10.

    I know that 3*7 = 21, but 32*7 = 224, too much, so I go with 6.
    6* 32 = 192, 217-192 =25.
    So 217 = 6*32 + 25.
    That means that 537 = 10*32 + 6*32 + 25.
    I can now write ##\frac{537}{32}= \frac{10*32}{32}+\frac{6*32}{32}+\frac{25}{32} = 16 + \frac{25}{32}##

    That is how long division works. If you don't enjoy the bookkeeping methods with the framework, then just break it up into fractions. Then you see all the pieces as you are putting them together.
     
  20. Dec 24, 2014 #19
    Been away from this thread for a while because I gave up. I'm going to try again.

    All that came up when I googled that was some crazy stuff like this:
    I don't understand any of that and hence I don't understand sequential approximation.
    I think this is similar to a set of videos I saw called the 'partial quotient method of division'. I am pretty sure I understand this but I don't understand how it explains the problem in the first post of the thread. For example, how would this work if it was 532 + 5 / 29 + 3?
     
  21. Dec 24, 2014 #20

    rcgldr

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    Polynomial division isn't the same as sequential approximation. Also using long hand division of decimal numbers as an example of polynomial division isn't very good either, because decimal long hand division involves borrows from higher order terms, and polynomial division does not involve borrows. It may help you to understand if you realize that only the most significant term of the divisor and the current divivdend is needed for each divide step, and that each quotient term is exact, not an approximation.

    Code (Text):

                           x + 2
                ________________
         x + 1  |   x^2 + 3x + 6

                    x^2            (bring down the x^2)
                                          (x^2 / x = x)
                 -( x^2 +  x )
                   ----------
                         -1x
                       +  3x       (bring down the  3x)
                        ----
                          2x
                                           (2x / x = 2)
                       -( 2x + 2)
                        ---------
                             - 2
                           +   6   (bring down the   6)
                             ---
                               4
     
     
    Last edited: Dec 24, 2014
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