# Factorisation of Monic Polynomials

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#2

(Original post by

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**rayquaza17**)...

E.g. in can be written as the product of 3 factors (2 linear, one quadratic), but it is actually divisible by 6 (7 if you include itself) different monic polynomials. If the quadratic/linear factors are a,b,c, then the polynomials factors are a,b,c,ab,bc,ac,abc, and are all different.

Apologies if this is of no use, or misleading. PM me if you wish me to delete it.

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#3

You can think of this in a similar way to the integers

If an integer is divisible by an integer then for some integer and so you've just found another integer that divides

That is to say, when you find integers that divide a number, you always find them in pairs... unless?

If an integer is divisible by an integer then for some integer and so you've just found another integer that divides

That is to say, when you find integers that divide a number, you always find them in pairs... unless?

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#4

(Original post by

You can think of this in a similar way to the integers

That is to say, when you find integers that divide a number, you always find them in pairs... unless?

**Noble.**)You can think of this in a similar way to the integers

That is to say, when you find integers that divide a number, you always find them in pairs... unless?

(Note, spoilers for OP past a certain point)

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#5

(although it's an identical argument anyway)

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#6

(Original post by

You can think of this in a similar way to the integers

If an integer is divisible by an integer then for some integer and so you've just found another integer that divides

That is to say, when you find integers that divide a number, you always find them in pairs... unless?

**Noble.**)You can think of this in a similar way to the integers

If an integer is divisible by an integer then for some integer and so you've just found another integer that divides

That is to say, when you find integers that divide a number, you always find them in pairs... unless?

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(Original post by

Not familiar with this, so can't really help, but I suspect a misunderstanding.

E.g. in can be written as the product of 3 factors (2 linear, one quadratic), but it is actually divisible by 6 (7 if you include itself) different monic polynomials. If the quadratic/linear factors are a,b,c, then the polynomials factors are a,b,c,ab,bc,ac,abc, and are all different.

Apologies if this is of no use, or misleading. PM me if you wish me to delete it.

**ghostwalker**)Not familiar with this, so can't really help, but I suspect a misunderstanding.

E.g. in can be written as the product of 3 factors (2 linear, one quadratic), but it is actually divisible by 6 (7 if you include itself) different monic polynomials. If the quadratic/linear factors are a,b,c, then the polynomials factors are a,b,c,ab,bc,ac,abc, and are all different.

Apologies if this is of no use, or misleading. PM me if you wish me to delete it.

(Original post by

Yes, that's what made me give the hint using integers

(although it's an identical argument anyway)

**Noble.**)Yes, that's what made me give the hint using integers

(although it's an identical argument anyway)

I can prove that if the number of divisors is odd, then the number itself is a square number (using unique prime factorisation):

Spoiler:

Show

So should I base this question on the proof for that?

If so, do you have any more hints?

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#9

(Original post by

Yeah that was helpful. I think it I was told r(x) had 9 roots then it would be okay to put what I said, I think I just got myself a bit confused!

We went through all of the maths behind the locker room problem (http://mathforum.org/alejandre/frisb...nt.locker.html) and I think that's the same as the video.

I can prove that if the number of divisors is odd, then the number itself is a square number (using unique prime factorisation):

So should I base this question on the proof for that?

If so, do you have any more hints?

**rayquaza17**)Yeah that was helpful. I think it I was told r(x) had 9 roots then it would be okay to put what I said, I think I just got myself a bit confused!

We went through all of the maths behind the locker room problem (http://mathforum.org/alejandre/frisb...nt.locker.html) and I think that's the same as the video.

I can prove that if the number of divisors is odd, then the number itself is a square number (using unique prime factorisation):

Spoiler:

Show

So should I base this question on the proof for that?

If so, do you have any more hints?

Look at the first eight unique monic polynomials that divide r(x), you must have

Now there's also a ninth monic polynomials that is different from the first eight, so

But there are only nine unique polynomials so for some

You can easily show that i cannot take values 1 to 8 (inclusive) because otherwise you'd have (WLOG)

with different, which clearly doesn't work (this is essentially saying, in the integers, if you divide a non-zero integer by two different integers you can get the same value).

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(Original post by

Yes, it's not really any different for polynomials.

Look at the first eight unique monic polynomials that divide r(x), you must have

Now there's also a ninth monic polynomials that is different from the first eight, so

But there are only nine unique polynomials so for some

You can easily show that i cannot take values 1 to 8 (inclusive) because otherwise you'd have (WLOG)

with different, which clearly doesn't work (this is essentially saying, in the integers, if you divide a non-zero integer by two different integers you can get the same value).

**Noble.**)Yes, it's not really any different for polynomials.

Look at the first eight unique monic polynomials that divide r(x), you must have

Now there's also a ninth monic polynomials that is different from the first eight, so

But there are only nine unique polynomials so for some

You can easily show that i cannot take values 1 to 8 (inclusive) because otherwise you'd have (WLOG)

with different, which clearly doesn't work (this is essentially saying, in the integers, if you divide a non-zero integer by two different integers you can get the same value).

I would never have come up with that in the exam though.

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#11

(Original post by

Right I get it now, thank you.

I would never have come up with that in the exam though.

**rayquaza17**)Right I get it now, thank you.

I would never have come up with that in the exam though.

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