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Aops intro to NT


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My kid has been taking this class and it has been really trouble free. This week he got everything but one problem done. He just can't solve it at all. When he asked for help on the message board, his teacher told him to use the long polynomial division. The trouble is it's not taught in Intro to Algebra. I though Intro to Number Theory was before Intermediate Algebra.

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That's a little odd to require long polynomial division in NT.  Perhaps it's just ONE easy approach?

 

Polynomial division isn't that tough to pick up, if needed (after Intro to Algebra).  DS picked it up in about 15 minutes this summer.  He hasn't done Intermediate yet.

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That's a little odd to require long polynomial division in NT. Perhaps it's just ONE easy approach?

 

Polynomial division isn't that tough to pick up, if needed (after Intro to Algebra). DS picked it up in about 15 minutes this summer. He hasn't done Intermediate yet.

Should I have him point out to the teacher that it isn't covered in what should be a prerequisite in the class?

There is no other way in that problem.

I will see if I have anything on the shelf that explains this.

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I think in AOPS that they throw out an occasionally really hard problem that they don't really expect many students to solve. It's likely that the instructor knows that polynomial division hasn't been covered but wants to see if anyone can figure it out. Is it possible that the section it is in is any any way related to polynomial division and could be a teaser for the kids of things to come? I would tell your DS it's ok to press the give up button. 

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I think in AOPS that they throw out an occasionally really hard problem that they don't really expect many students to solve. It's likely that the instructor knows that polynomial division hasn't been covered but wants to see if anyone can figure it out. Is it possible that the section it is in is any any way related to polynomial division and could be a teaser for the kids of things to come? I would tell your DS it's ok to press the give up button.

It's in divisor arithmetic, otherwise an easy section. All that's coming our way is modular arithmetic, which I know absolutely nothing about.

Funny thing is even if he gives up, his bar for homework problems will still be blue, but this kid won't push that button. I have got to figure out if there is a khan video or anything that could teach him the trick for now.

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That's a little odd to require long polynomial division in NT. Perhaps it's just ONE easy approach?

 

Polynomial division isn't that tough to pick up, if needed (after Intro to Algebra). DS picked it up in about 15 minutes this summer. He hasn't done Intermediate yet.

You are right. It took the first 6 minutes of a 12 minute Khan video to understand it at a level he needed. 😂

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I think it's just one possible solution technique, and not essential for the topic. Sometimes the solution writer could be a different person to the question writer, and they might not have the same solution idea in mind. Also the solution writer might not have a precise idea of what the student is "supposed to know".

 

Anyway, it's worthwhile knowing polynomial long division, and if you google it you quickly see many links, and the able student can get it.

 

By the way, if you happen to be dividing polynomial p(x) by linear function x-a (here x is the variable, and a is the constant), then the remainder is simply p(a), i.e. plug in a for x in p(x) to get the remainder!

 

Edited by epi
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I think it's just one possible solution technique, and not essential for the topic. Sometimes the solution writer could be a different person to the question writer, and they might not have the same solution idea in mind. Also the solution writer might not have a precise idea of what the student is "supposed to know".

 

Anyway, it's worthwhile knowing polynomial long division, and if you google it you quickly see many links, and the able student can get it.

 

By the way, if you happen to be dividing polynomial p(x) by linear function x-a (here x is the variable, and a is the constant), then the remainder is simply p(a), i.e. plug in a for x in p(x) to get the remainder!

Actually in this case you were dividing a quadratic on x+23, so the only way to solve it was to use the long polynomial division. Interestingly the problem itself was super easy once the division was done, of course. 😂 And there was no way of solving it without it because the solution was to be found using a remainder.

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i used to be afraid of polynomial division until i realized it is just like integer division only easier. i.e. an integer is essentially a polynomial with x set equal to 10, only there is no carrying needed for polynomials, hence your first guess in long division always works!  e.g. when dividing 649 by 38,  our first guess of 2 (actually 20), does not work because of carrying, but if we divide 6X^2 + 4X + 9 by 3X + 8, then the first guess of 2X does work!  I think I myself was a college professor teaching this stuff by the time i noticed this, but probably some of your kids could figure this out on their own.

 

(when i speak of "carrying" or not, of course i mean that with integers, a large enough number of ones changes the number of tens, and enough tens changes the number of hundreds, but with polynomials, no number of ones changes the number of X's and no number of X's changes the number of X^2's.)

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as for modular artihmetic, you are just setting multiples of some fixed integer, the modulus, equal to zero, so the result is the remainder after dividing by the modulus.  this sort of thing was in old days called "casting out", i.e. "casting out nines" meant setting multiples of 9 equal to zero and seeing what was left.  this is easy since in base 10, every power of 10 gives 1 when nines are cast out, so 638 becoms 6+3+8 = 9+8 = 8, after casting out nines.  casting out 10's is the easiest since it just leaves the ones term, i.e. 638 is equal to 8 "modulo 10".

 

you can cast out multiples of polynomials too, and then a polynomial p(X) is equivalent modulo X-a to what ever is the remainder after division by X-a.  Of course this means multiples of X-a are set equal to zero, in particular X-a is set equal to zero, so X is set equal to a, which means of course the result, i.e. the remainder, is just p(a).

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